Explanation:
First lets understand the 2nd law of motion by Sir Isaac Newton. According to this law,
[tex]Force = mass\times acceleration[/tex]
[tex]Velocity = acceleration\times time[/tex]
Since the spy is in space there is no medium and hence no friction to restrict the motion. Thus, when the spy turns on the jet pack, she will be accelerated and her velocity will increase. As there is no medium so no friction. So even when she turns her jet-pack off the velocity will not change. Although the acceleration will be zero but she will be moving with a constant velocity until an opposite force is applied. That can be done using reverse thrust.
A student pushes a 50 kg box of books on a flat surface with a force of 120 N at an angle of 60° on the horizontal. If the surface is friction free, how far does the box move in 5 seconds assuming it started moving from rest? a. 6.0 m
b. 8.5 m
c. 15 m
d. 4.6 m
Answer:
c. 15 m
Explanation:
We apply Newton's second law in the x direction:
∑Fₓ = m*a
120*cos(60°) = 50*a
[tex]a = \frac{120*cos(60^o)}{50} = 1.2 \frac{m}{s^2}[/tex]
Block kinematics
The block moves with uniformly accelerated movement, so we apply the following formula to calculate the distance
[tex]d = V_o*t + \frac{1}{2}*a*t^2[/tex]
[tex]d = 0 + \frac{1}{2}*1.2*5^2[/tex]
d = 15m
Drying of Cassava (Tapioca) Root. Tapioca flour is used in many countries for bread and similar products. The flour is made by drying coarse granules of the cassava root containing 66 wt % moisture to 5% moisture and then grinding to produce a flour. How many kg of granules must be dried and how much water removed to produce 5000 kg/h of flour?
To produce 5000 kg/h of tapioca flour with 5% moisture from cassava granules with 66% moisture, 13970.59 kg of granules must be dried, resulting in 8966.59 kg of water being removed.
The question pertains to the process of drying cassava root to produce tapioca flour, which involves reducing the moisture content from 66 wt % to 5%. To find the weight of cassava granules needed to produce 5000 kg/h of flour, we utilise mass balance concepts.
Let x be the amount (kg) of granules required. These granules initially contain 66% moisture, so there are 0.34x kg of dry solids in them. After drying to 5% moisture, the 5000 kg of flour contains 95% dry solids, or 0.95 x 5000 kg.
Assuming no loss of solid material during drying:
0.34x = 0.95 x 5000
x = (0.95 x 5000) / 0.34
x ≈ 13970.59 kg
The initial weight of water in the granules is the total weight of granules minus the weight of dry solids:
Initial water weight = x - 0.34x
Initial water weight = 0.66x
Initial water weight = 0.66 x 13970.59 kg
Initial water weight ≈ 9216.59 kg
The final weight of water in the 5000 kg of flour at 5% moisture is:
Final water weight = 0.05 x 5000 kg
Final water weight = 250 kg
The amount of water removed during the drying process is the initial water weight minus the final water weight.
Water removed = 9216.59 kg - 250 kg
Water removed ≈ 8966.59 kg
A vector has an x-component of -26.5 and a y-component of 43 units. Find magnitude and direction of vector
Answer:
The vector has a magnitude of 33.86 units and a direction of 121.64°.
Explanation:
To find the magnitude, you use the Pitagorean theorem:
[tex]||V|| = \sqrt{x^2 + y^2}= \sqrt{(-26.5)^2 + (43)^2} = 33.86 units[/tex]
In order to find the direction, you can use trigonometry. You have to keep in mind, that as the y component of the vector is positive and the x component is negative, the vector must have an angle between 90 and 180°, or in the second quadrant of the plane.:
[tex]tan(\alpha) = \frac{y}{x} \\\alpha = tan^{-1}(\frac{43}{-26.5})= 180 - tan^{-1}(\frac{43}{26.5}) = 121.64[/tex]° or, 58.36° in the second quadrant
A quarter circle of radius a is centered about the origin in the first quadrant and carries a uniform charge of −Q. Find the x- and y-components of the net electric field at the origin.
Answer:
[tex]E_x = \frac{2kQ}{\pi R^2}[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2}[/tex]
Explanation:
Electric field due to small part of the circle is given as
[tex]dE = \frac{kdq}{R^2}[/tex]
here we know that
[tex]dq = \frac{Q}{\frac{\pi}{2}R} Rd\theta[/tex]
[tex]dq = \frac{2Q d\theta}{\pi}[/tex]
Now we will have two components of electric field given as
[tex]E_x = \int dE cos\theta[/tex]
[tex]E_x = \int \frac{kdq}{R^2} cos\theta[/tex]
[tex]E_x = \int \frac{k (2Qd\theta) cos\theta}{\pi R^2}[/tex]
[tex]E_x = \frac{2kQ}{\pi R^2} \int_0^{90} cos\theta d\theta[/tex]
[tex]E_x = \frac{2kQ}{\pi R^2} (sin 90 - sin 0)[/tex]
[tex]E_x = \frac{2kQ}{\pi R^2}[/tex]
similarly in Y direction we have
[tex]E_y = \int dE sin\theta[/tex]
[tex]E_y = \int \frac{kdq}{R^2} sin\theta[/tex]
[tex]E_y = \int \frac{k (2Qd\theta) sin\theta}{\pi R^2}[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2} \int_0^{90} sin\theta d\theta[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2} (-cos 90 + cos 0)[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2}[/tex]
You have two square metal plates with side length of 16.50 cm. You want to make a parallel-plate capacitor that will hold a charge of 18.5 nC when connected to a 37.8 V potential difference. Determine the necessary separation in mm. Round your answer to three significant figures.
Answer:
d = 3.44 *10^{-7} m
Explanation:
given data:
length of metal plates = 16.50 cm
capacitor charge = 18.5 nC
potential difference = 37.8 V
capacitance of parallel plate capacitor
[tex]C = \frac{A\epsilon _{0}}{d}[/tex]
area of the individual plate
A=[tex] a^2 = (16.5*10^{-2})^2 = 272.25 *10^{-4}[/tex] m2
capacitance
[tex]C = QV = 18.5 *10^{-9} *37.8 = 699.3 * 10^{-9} C[/tex]
separation between plates d is given as[tex] = \frac{A\epsilon _{0}}{C }[/tex]
[tex]d = \frac{272.25 *10^{-4} *8.85*10^{-12}}{699.3 *10^-9}[/tex]
d = 3.44 *10^{-7} m
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC (nano-coulombs) is placed into this region. What is the y component of the electric force on this charge?
Answer:
[tex]-1.144\ \mu C[/tex]
Explanation:
Given:
[tex]\vec{E}[/tex] = uniform electric field in the space = [tex](148.0\ \hat{i}-110.0\ \hat{j})\ N/C[/tex]Q = Charge placed in the region = [tex]10.4 nC\ = 1.04\times 10^{-8}\ nC[/tex]Assume:
[tex]\vec{F}[/tex] = Electric force on the charge due to electric fieldWe know that the electric field is the electric force applied on a unit positive charge i.e.,
[tex]\vec{E}=\dfrac{\vec{F}}{Q}[/tex]
This means the electric force applied on this additional charge placed in the field is given by:
[tex]\vec{F}=Q\vec{E}\\\Rightarrow \vec{F} = 1.04\times 10^{-8}\ n C\times (148.0\ \hat{i}-110.0\ \hat{j})\ N/C\\\Rightarrow \vec{F} = (1.539\ \hat{i}-1.144\ \hat{j})\ \mu N\\[/tex]
From the above expression of force, we have the following y-component of force on this additional charge.
[tex]F_y = -1.144\ \mu N[/tex]
Hence, the y-component of the electric force on the this charge is [tex]-1.144\ \mu N[/tex].
David Scott's experiment shows that all objects, regardless of their weight, fall equally in the absence of: a. The gravity
b. The air
c. The pressure
d. The gravitational force
Answer:
b. the air
Explanation:
David Scott's experiment was performed on the moon, where there is gravity but there is no air. This experiment consisted of letting a hammer and a feather fall at the same time. The result was that the two objects touch the ground simultaneously.
Since these two objects obviously have a different mass, the experiment shows that in a vacuum, objects fall with the same acceleration regardless of their mass.
A person is riding on a Ferris Wheel. When the wheel makes one complete turn, is the net work done by the gravitational force posiitive, negative, or zero?
Answer:
Zero
Explanation:
The overall work done by gravitational force on completion of a complete turn is zero.
Since the work done by gravitaional force is conservative and depends only on the initial and end position here height and the path followed does not matter.
Since, in a complete turn the wheel return to its final position as a result of which displacement is zero and as work is the dot product of Force exerted and displacement, the work done is zero.
Also the work done in half cycle by gravity is counter balance by the work which is done against the gravity in the other half cycle.
The net work done by the gravitational force when a person is riding on a Ferris Wheel and it makes one complete turn is zero.
To understand why the net work done by the gravitational force is zero, we need to consider the definition of work and the nature of the motion on a Ferris Wheel. Work done by a force is defined as the product of the force and the displacement in the direction of the force. Mathematically, this is expressed as:
[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]
where ( W ) is the work, ( F ) is the force, ( d ) is the displacement, and [tex]\( \theta \)[/tex] is the angle between the force and the displacement.
In the case of the Ferris Wheel, the gravitational force acts vertically downward towards the center of the Earth, while the displacement of the person on the Ferris Wheel is along the circumference of the wheel, which is horizontal at any given point. Since the force and displacement are perpendicular to each other at every point in the circle, the angle \( \theta \) between them is always 90 degrees. Therefore, the cosine of 90 degrees is zero, which means that the work done by the gravitational force at each point is zero:
[tex]\[ W = F \cdot d \cdot \cos(90^\circ) = F \cdot d \cdot 0 = 0 \][/tex]
Moreover, over one complete turn of the Ferris Wheel, the initial and final positions of the person are the same. This means that the total displacement over one complete cycle is zero. Even if we consider the components of the gravitational force along the direction of displacement during different parts of the cycle, the net effect is zero because the person is raised to a certain height and then lowered back to the starting point. The work done against gravity to raise the person is equal in magnitude and opposite in sign to the work done by gravity as the person descends, resulting in a net work of zero for the entire cycle.
Therefore, the gravitational force does no net work on the person over one complete turn of the Ferris Wheel.
A cheetah can accelerate from rest to a speed of 21.5 m/s in 6.75 s. What is its acceleration? m/s^2
Answer:
Acceleration will be [tex]a=3.185m/sec^2[/tex]
Explanation:
We have given final velocity v = 21.5 m/sec
Time t = 6.75 sec
As cheetah starts from rest so initial velocity u = 0 m/sec
From first equation of motion we know that v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time
So [tex]21.5=0+a\times 6.75[/tex]
[tex]a=3.185m/sec^2[/tex]
Answer:
[tex]a=3.185\frac{m}{s^2}[/tex]
Explanation:
Acceleration is the change in velocity for a given period of time, we can express this in the next formula:
[tex]a = \frac{\Delta v}{\Delta t} =\frac{v_{1}-v_{0}}{t_{1}-t_{0}}[/tex]
In this case the values are:
[tex]v_{0}=0\\v_{1}= 21.5 m/s\\t_{0}=0\\t_{1}= 6.75 s\\[/tex]
Inserting known values, the acceleration is:
[tex]a= \frac{21.5 m/s}{6.75 s} \\a=3.185\frac{m}{s^2}[/tex]
On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.00 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 44.4 m/s at an angle of 25° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are (a) the maximum height and (b) the range of the ball?
Answer:
(a) Maximum height = 53.88 meters
(b) Range of the ball = 924.36 meters
Explanation:
The ball has been launched at a speed = 44.4 meters per second
Angle of the ball with the horizontal = 25°
Horizontal component of the speed of the ball = 44.4cos25° = 40.24 meters per second
Vertical component = 44.4sin25° = 18.76 meters per second
We know vertical component of the speed decides the height of the ball so by the law of motion,
v² = u² - 2gh
where v = velocity at the maximum height = 0
u = initial velocity = 18.76 meter per second
g = gravitational force = 9.8 meter per second²
Now we plug in the values in the given equation
0 = (18.76)² - 2(9.8)(h)
19.6h = 352.10
h = [tex]\frac{352.10}{19.6}[/tex]
h = 17.96 meters
By another equation,
[tex]v=ut-\frac{1}{2}gt^{2}[/tex]
Now we plug in the values again
[tex]0=(18.76)t-\frac{1}{2}(9.8)t^{2}[/tex]
18.76t = 4.9t²
18.76 = 4.9t
t = [tex]\frac{18.76}{4.9}=3.83[/tex]seconds
Since time t is the time to cover half of the range.
Therefore, time taken by the ball to cover the complete range = 2×3.83 = 7.66 seconds
So the range of the ball = Horizontal component of the velocity × time
= 40.24 × 7.66
= 308.12 meters
This we have calculated all for our planet.
Now we take other planet.
(a) Since the golfer drives the ball 3 times as far as he would have on earth then maximum height achieved by the ball = 17.96 × 3 = 53.88 meters
(b) Range of the ball = 3×308.12 = 924.36 meters
How fast does a 2 MeV fission neutron travel through a reactor core?
Answer:
The answer is [tex] 1.956 \times 10^7\ m/s[/tex]
Explanation:
The amount of energy is not enough to apply the relativistic formula of energy [tex]E = mc^2[/tex], so the definition of energy in this case is
[tex]E = \frac{1}{2}m v^2[/tex].
From the last equation,
[tex]v= \sqrt{2E/m}[/tex]
where
[tex]E = 2 MeV = 3.204 \times 10^{-13} J[/tex]
and the mass of the neutron is
[tex]m = 1.675\times 10^{-27}\ Kg[/tex].
Then
[tex]v = 1.956 \times 10^7\ m/s[/tex]
the equivalent of [tex]0.065[/tex] the speed of light.
The potential difference between A and B is 5.0 V. A proton starts from rest at A. When it reaches B what is its kinetic energy? (e = 1.60 x 10^-19 C)
Answer:
total kinetic energy is 8 × [tex]10^{-19}[/tex] J
Explanation:
given data
potential difference = 5 V
e = 1.60 × [tex]10^{-19}[/tex] C
to find out
what is kinetic energy
solution
we will apply here conservation of energy that is
change in potential energy is equal to change in kinetic energy
so
change potential energy is e × potential difference
change potential energy = 1.60 × [tex]10^{-19}[/tex] × 5
change potential energy = 8 × [tex]10^{-19}[/tex] J
so change in kinetic energy = 8 × [tex]10^{-19}[/tex] J
and we know proton start from rest that mean ( kinetic energy is 0 ) so
change in KE is total KE
total kinetic energy is 8 × [tex]10^{-19}[/tex] J
Three charges, each of magnitude 10 nC, are at separate corners of a square of edge length 3 cm. The two charges at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a fourth charge q = +3 nC at the remaining (upper right) corner. (Assume the +x axis is directed to the right and the +y axis is directed upward.)
Answer:
The force exerted by three charges on the fourth is [tex]F_{resultant}=2.74\times10^{-5}\ \rm N[/tex]
Explanation:
Given:
The magnitude of three identical charges, [tex]q=10\ \rm nC[/tex]Length of the edge of the square a=3 cmMagnitude of fourth charge ,Q=3 nCAccording to coulombs Law the force F between any two charge particles is given by
[tex]F=\dfrac{kQq}{r^2}[/tex]
where r is the radial distance between them.
Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition
[tex]F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N[/tex]
The force on the fourth charge is calculated by first determining the individual forces exerted by each of the three other charges separately using Coulomb's Law and then adding these forces as vectors. This involves resolving each force into its x and y components, combining them separately, and then determining the resultant force's magnitude and direction.
Explanation:The problem here involves Coulomb's Law and the superposition principle in physics. Coulomb's Law defines the force between two point charges as directly proportional to the product of their charges, and inversely proportional to the square of the distance between them.
First, you need to calculate the forces exerted on the fourth charge by each of the three other charges separately. This involves calculating the distance from each existing charge to the fourth charge, then subbing these distances, along with the relevant charge values, into the Coulomb's Law formula. Remember that if the charge is positive (like in the case of charge +q), the force vector points directly from the charge, while if the charge is negative, the force vector points towards the charge.
After calculating the force vectors resulting from each charge, you add these vectors together to get the resultant force vector which is the force exerted on the fourth charge. This problem also involves trigonometry as when you add the force vectors, you have to take into account the direction which each force vector is pointing.
Force due to the positive charge at the lower left: F1 is in the first quadrant
Force due to the positive charge at the lower right: F2 is in the fourth quadrant
Force due to the negative charge at the upper left: F3 is in the third quadrant
In each case, you'll need to resolve each force into its x and y components, and then add up all the x and y components separately to get the x and y components of the total force. Finally, calculate the magnitude of the total force using the Pythagorean theorem.
Learn more about Coulomb's Law here:https://brainly.com/question/506926
#SPJ3
An arrow is shot from a height of 1.7 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.4 s later. What is the height of the cliff?
what is the maximum height (in m) reached by the arrow along it trajectory?
what is the arrows impact speed (in m/s) just before hitting the cliff?
The height of the cliff is 16.97 m, the maximum height reached by the arrow is 24.47 m, and the impact speed of the arrow just before hitting the cliff is 16.47 m/s.
Explanation:In this scenario, we can apply the equations of motion to calculate the height of the cliff, the maximum height reached by the arrow, and its impact speed.
Firstly, the height of the cliff can be calculated using the equation Y = Yo + Vy*t - 0.5*g*t^2, where g is the gravity, t is the time, Vy is the initial vertical speed, and Yo is the initial height. Given Yo = 1.7m, Vy = 26sin(60°), t = 3.4s, and g = 9.8 m/s^2, the height H of the cliff is 16.97 m.
Secondly, the maximum height reached by the arrow can be calculated by the equation Hmax = Yo + Vy*t - 0.5*g*(t)^2, where t is the time it takes to reach the maximum height, which can be Ve/g. Ve is the initial vertical velocity whose value is Vy = 26sin(60°). Hence the maximum height Hmax is 24.47 m.
Finally, the arrow’s impact speed can be calculated by using Pythagoras' theorem. The impact speed V = sqrt((Vx)^2 + (Vy)^2), where Vx is the horizontal velocity and Vy is the final vertical velocity. Given Vx = 26cos(60°) and Vy = Ve - g*t, with Ve = 26sin(60°) and t = 3.4s, the impact speed V of the arrow just before hitting the cliff is 16.47 m/s.
Learn more about Projectile Motion here:https://brainly.com/question/29545516
#SPJ12
The height of the cliff is 21.6 m, and the arrow’s impact speed just before hitting the cliff is approximately 16.8 m/s. These calculations use projectile motion equations for both components.
A. To determine the height of the cliff, we can use the vertical motion equation:
[tex]y = y_0 + v_{0y}t - 0.5gt^2[/tex]
Where:
y₀ = initial height = 1.7 m[tex]v_{0y}[/tex] = initial vertical velocity = v₀sin(θ) = 26sin(60°) = 22.5 m/sg = acceleration due to gravity = 9.8 m/s²t = time = 3.4 sSubstituting these values into the equation:
y = 1.7 + 22.5(3.4) - 0.5(9.8)(3.4)2
y = 1.7 + 76.5 - 56.6 = 21.6 m
Therefore, the height of the cliff is 21.6 m.
B. To find the impact speed, we need to calculate both the final vertical and horizontal components of velocity:
Horizontal component (vₓ): It remains constant:The total impact speed is found using the Pythagorean theorem:
[tex]v_f = \sqrt{(v_x^2 + v_{fy}^2)} = \sqrt{(132 + (-10.8)2)} \approx 16.8 m/s[/tex]
Thus, the arrow's impact speed is approximately 16.8 m/s.
A movie star catches a paparazzi reporter snapping pictures of her at home and claims that he was trespassing. He, of course denies the allegations. To prove her point, she submits as evidence the film that she confiscated. Her height of 1.75 m appears as an 8.25 mm high image on the film. Additionally, she submits that the camera that was used has a focal length of 210 mm. How far away was the reporter when he took the picture? (All the information that is given).
Answer:
44.755 m
Explanation:
Given:
Height of the movie star, H = 1.75 m = 1750 mm
Height of the image, h = - 8.25 mm
Focal length of the camera = 210 mm
Let the distance of the object i.e the distance between camera and the movie star be 'u'
and
distance between the camera focus and image be 'v'
thus,
magnification, m = [tex]\frac{\textup{h}}{\textup{H}}[/tex]
also,
m = [tex]\frac{\textup{-v}}{\textup{u}}[/tex]
thus,
[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{h}}{\textup{H}}[/tex]
or
[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{-8.25}}{\textup{1750}}[/tex]
or
[tex]\frac{\textup{1}}{\textup{v}}=-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex] ....................(1)
now, from the lens formula
[tex]\frac{\textup{1}}{\textup{f}}=\frac{\textup{1}}{\textup{u}}+\frac{1}{\textup{v}}[/tex]
on substituting value from (1)
[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}+-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex]
or
[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}(1 -\frac{\textup{1750}}{\textup{-8.25}})[/tex]
or
u = 210 × ( 1 + 212.12 )
or
u = 44755.45 mm
or
u = 44.755 m
Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 cm . Charge q1 is at x = 2.50 cm .What is q1 (magnitude and sign) if the net force on q3 is zero?
Answer:
q₁= +0.5nC
Explanation:
Theory of electrical forces
Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
o solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters
Data:
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Data
q₃=+5.00 nC =+5* 10⁻⁹ C
q₂= -2.00 nC =-2* 10⁻⁹ C
d₂= 5.00 cm= 5*10⁻² m
d₁= 2.50 cm= 2.5*10⁻² m
k = 8.99*10⁹ N*m²/C²
Calculation of magnitude and sign of q1
Fn₃=0 : net force on q3 equals zero
F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.
F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.
We propose the algebraic sum of the forces on q₃
F₂₃ - F₁₃=0
[tex]\frac{k*q_{2} *q_{3} }{d_{2}^{2} } -\frac{k*q_{1} *q_{3} }{d_{1}^{2} }=0[/tex]
We eliminate k*q₃ of the equation
[tex]\frac{q_{1} }{d_{1}^{2} } = \frac{q_{2} }{d_{2}^{2} }[/tex]
[tex]q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2} }[/tex]
[tex]q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4} }{5^{2}*10^{-4} }[/tex]
q₁= +0.5*10⁻⁹ C
q₁= +0.5nC
Final answer:
The magnitude and sign of charge q1 can be found by setting the electric force by q1 on q3 equal and opposite to the force exerted by q2 on q3 using Coulomb's Law and solving for q1 given the known distances.
Explanation:
The question is asking to find the magnitude and sign of charge q1 such that the net electrostatic force on charge q3 is zero when placed in a line with charges q2 and q3. To solve this, we can apply Coulomb's law, which states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
To have the net force on q3 be zero, the force exerted on q3 by q1 must be equal in magnitude and opposite in direction to the force exerted on q3 by q2. We can set up equations based on Coulomb's law and solve for q1. First, determine the force F31 between q3 and q1, and the force F32 between q3 and q2. Since these forces must be equal and opposite to cancel each other out, we set F31 = F32 and solve for q1. Since the distance between q3 and q1 is 2.50 cm and between q3 and q2 is 5.00 cm, and accounting for the sign of q2, we can calculate the magnitude and sign of q1 that makes the net force on q3 zero.
What are the (time varying) amplitudes of the E and H fields if summer sunlight has an intensity of 1150 W/m2 in any Town?
Calculate the relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth.
Answer:
The relative strength of the gravitational and solar electromagnetic pressure forces is [tex]7.33\times10^{13}\ N[/tex]
Explanation:
Given that,
Intensity = 1150 W/m²
(a). We need to calculate the magnetic field
Using formula of intensity
[tex]I=\dfrac{E^2}{2\mu_{0}c}[/tex]
[tex]E=\sqrt{2\times I\times\mu_{0}c}[/tex]
Put the value into the formula
[tex]E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}[/tex]
[tex]E=931.17\ N/C[/tex]
Using relation of magnetic field and electric field
[tex]B=\dfrac{E}{c}[/tex]
Put the value into the formula
[tex]B=\dfrac{931.17}{3\times10^{8}}[/tex]
[tex]B=0.0000031039\ T[/tex]
[tex]B=3.10\times10^{-6}\ T[/tex]
(2). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth
We need to calculate the gravitational force
Using formula of gravitational
[tex]F_{g}=\dfrac{GmM}{r^2}[/tex]
Where, m = mass of sun
m = mass of earth
r = distance
Put the value into the formula
[tex]F_{g}=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}[/tex]
[tex]F_{g}=3.52\times10^{22}\ N[/tex]
We need to calculate the radiation force
Using formula of radiation force
[tex]F_{R}=\dfrac{I}{c}\times\pi\timesR_{e}^2[/tex]
[tex]F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.371\times10^{6})^2[/tex]
[tex]F_{R}=4.8\times10^{8}\ N[/tex]
We need to calculate the pressure
[tex]\dfrac{F_{g}}{F_{R}}=\dfrac{3.52\times10^{22}}{4.8\times10^{8}}[/tex]
[tex]\dfrac{F_{g}}{F_{R}}=7.33\times10^{13}\ N[/tex]
Hence, The relative strength of the gravitational and solar electromagnetic pressure forces is [tex]7.33\times10^{13}\ N[/tex]
Baseball homerun hitters like to play in Denver, but
curveballpitchers do not. Why?
Answer:
Because of height and lower atmospheric pressure.
Explanation:
Atmospheric pressure affects aerodynamic drag, lower pressure means less drag. At the altitude of Denver the air has lower pressure, this allows baseball players to hit balls further away.
Another aerodynamic effect is the Magnus effect. This effect causes spinning objects to curve their flightpath, which is what curveball pitchers do. A lower atmospheric pressure decreases the curving of the ball's trajectory.
Denver's high altitude results in lower air pressure which benefits homerun hitters as the baseball can travel further. However, this is disadvantageous for curveball pitchers as the lesser air pressure makes it harder to produce a good curve.
Explanation:Baseball home run hitters and curveball pitchers react differently to playing in Denver. Denver is located at a high altitude, which means the air pressure is lower than in many other cities. A lower air pressure means there’s less air resistance. For hitters, less air resistance means that the baseball can travel further when hit, increasing the likelihood of hitting a home run.
However, for pitchers who throw curveballs, the low air pressure is not beneficial. This is because the curve of a curveball is produced by the difference in air pressure on either side of the ball. Notably, the spin that the pitcher puts on the ball makes the air pressure higher on one side of the ball and lower on the other. However, the reduced air density in Denver reduces the overall air pressure difference, making it harder to get good curves on their pitches. Thus, hitters like to play in Denver while pitchers prefer places with denser air.
Learn more about the Effects of Air Pressure on Baseball here:https://brainly.com/question/14243397
#SPJ12
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an angle of θ = 30◦ to the ground. The coefficient of kinetic friction μk = 0.2. At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction μk = 0.3 until it comes to rest. The goal of this problem is to find out how far the object slides along the rough surface. What is the work done by the friction force while the mass is sliding down the in- clined plane? (Is it positive or negative?) (b) What is the work done by the gravitational force while the mass is sliding down the inclined plane? (Is it positive or negative?)
Answer:
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:
Wg= 58.8 J is positive
Explanation:
Nomenclature
vf: final velocity
v₀ :initial velocity
a: acceleleration
d: distance
Ff: Friction force
W: weight
m:mass
g: acceleration due to gravity
Graphic attached
The attached graph describes the variables related to the kinetics of the object (forces and accelerations)
Calculation de of the components of W in the inclined plane
W=m*g
Wx₁ = m*g*sin30°
Wy₁= m*g*cos30°
Object kinematics on the inclined plane
vf₁²=v₀₁²+2*a₁*d₁
v₀₁=0
vf₁²=2*a₁*d₁
[tex]v_{f1} = \sqrt{2*a_{1}*d_{1} }[/tex] Equation (1)
Object kinetics on the inclined plane (μ= 0.2)
∑Fx₁=ma₁ :Newton's second law
-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁
-μ₁N₁+Wx₁ = ma₁ Equation (2)
∑Fy₁=0 : Newton's first law
N₁-Wy₁= 0
N₁- m*g*cos30°=0
N₁ = m*g*cos30°
We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)
-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m
-μ₁*g*cos30°+g*sin30° = a₁
g*(-μ₁*cos30°+sin30°) = a₁
a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²
We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)
[tex]v_{f1} = \sqrt{2*3.2*3} }[/tex]
[tex]v_{f1} =\sqrt{2*3.2*3}[/tex]
[tex]v_{f1} = 4.38 m/s[/tex]
Rough surface kinematics
vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s
0 =4.38²+2*a₂*d₂ Equation (3)
Rough surface kinetics (μ= 0.3)
∑Fx₂=ma₂ :Newton's second law
-Ff₂=ma₂
--μ₂*N₂ = ma₂ Equation (4)
∑Fy₂= 0 :Newton's first law
N₂-W=0
N₂=W=m*g
We replace N₂=m*g inthe equation (4)
--μ₂*m*g = ma₂ We divide by m
--μ₂*g = a₂
a₂ =-0.2*9.8= -1.96m/s²
We replace a₂ = -1.96m/s² in the equation (3)
0 =4.38²+2*-1.96*d₂
3.92*d₂ = 4.38²
d₂=4.38²/3.92
d₂=4.38²/3.92
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf = - Ff₁*d₁
Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N
Wf= - 6.79*3 = 20.4 N*m
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane
Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m
Wg= 58.8 J is positive
The work done by the friction force is negative, and the work done by the gravitational force is positive as the object slides down the inclined plane.
Explanation:The work done by the friction force while the object slides down the inclined plane is negative. The work done by the gravitational force while the object slides down the inclined plane is positive.
When the object slides down the inclined plane, the friction force acts in the opposite direction to its motion. Since friction always opposes the motion, the work done by friction is negative.
The gravitational force, on the other hand, acts in the same direction as the object's motion. Therefore, the work done by the gravitational force is positive.
Learn more about work done by friction and gravitational force here:https://brainly.com/question/34111949
#SPJ3
A certain car takes 30m to stop when it is traveling at 25m/s. If a pedestrian is 28m in front of this car when the driver starts braking (starting at 25m/s), how long does the pedestrian have to get out of the way?
Answer:
It take 1.78033 second get away
Explanation:
We have given that a car takes 30 m to stop when its speed is 25 m/sec
As the car stops its final speed v = 0 m/sec
Initial speed u = 25 m/sec
Distance s = 30 m
From third law of motion [tex]v^2=u^2+2as[/tex]
So [tex]0^2=25^2+2\times a\times 30[/tex]
[tex]a=-10.4166m/sec^2[/tex]
Now in second case distance s = 28 m
So [tex]v^2=25^2+2\times -10.4166\times 28[/tex]
[tex]v^2=41.666[/tex]
v = 6.4549 m/sec
Now from first equation of motion v=u+at
So [tex]6.4549=25-10.4166\times t[/tex]
t = 1.78033 sec
An arctic weather balloon is filled with 12.1L of helium gas inside a prep shed. The temperature inside the shed is 9.°C. The balloon is then taken outside, where the temperature is −7.°C. Calculate the new volume of the balloon. You may assume the pressure on the balloon stays constant at exactly 1atm. Round your answer to 3 significant digits.
Answer : The new or final volume of gas will be, 11.4 L
Explanation :
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 12.1 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]9^oC=273+9=282K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]-7^oC=273+(-7)=266K[/tex]
Now put all the given values in the above formula, we get the final volume of the gas.
[tex]\frac{12.1L}{V_2}=\frac{282K}{266K}[/tex]
[tex]V_2=11.4L[/tex]
Therefore, the new or final volume of gas will be, 11.4 L
Final answer:
Using Charles's Law, the new volume of the helium gas in the balloon when taken from an inside temperature of 9°C to an outside temperature of -7°C, at constant pressure, is calculated to be 11.4 L.
Explanation:
To calculate the new volume of the helium gas in the balloon when it is taken outside to a colder temperature, we can use Charles's Law, which states that for a given mass of gas at constant pressure, the volume is directly proportional to its temperature in kelvins (V/T = k). We need to convert the temperatures from Celsius to Kelvin (K = °C + 273.15) and then apply Charles's Law (V1/T1 = V2/T2).
First, convert the temperature from Celsius to Kelvin:
Inside temperature: T1 = 9 °C + 273.15 = 282.15 K
Outside temperature: T2 = -7 °C + 273.15 = 266.15 K
Next, apply Charles's Law to find the new volume (V2):
V1/T1 = V2/T2
Plugging in the known values:
12.1 L / 282.15 K = V2 / 266.15 K
Solving for V2, we get:
V2 = (12.1 L × 266.15 K) / 282.15 K
V2 = 11.4 L (rounded to three significant digits)
Therefore, the new volume of the balloon when taken outside will be 11.4 L.
Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric field when there are two fields, E1 and E2, where E1 is directed vertically upward and has magnitude 99 V/m and E2 is directed 48° to the left of E1 and has magnitude 164 V/m.
Answer:
The resultant field will have a magnitude of 241.71 V/m, 30.28° to the left of E1.
Explanation:
To find the resultant electric fields, you simply need to add the vectors representing both electric field E1 and electric field E2. You can do this by using the component method, where you add the x-component and y-component of each vector:
E1 = 99 V/m, 0° from the y-axis
E1x = 0 V/m
E1y = 99 V/m, up
E2 = 164 V/m, 48° from y-axis
E2x = 164*sin(48°) V/m, to the left
E2y = 164*cos(48°) V/m, up
[tex]Ex: E_{1_{x}} + E_{2_{x}} = 0 V/m - 164 *sin(48) V/m= -121.875 V/m\\Ey: E_{1_{y}} + E_{2_{y}} = 99 V/m + 164 *cos(48) V/m = 208.74 V/m\\[/tex]
To find the magnitude of the resultant vector, we use the pythagorean theorem. To find the direction, we use trigonometry.
[tex]E_r = \sqrt{E_x^2 + E_y^2}= \sqrt{(-121.875V/m)^2 + (208.74V/m)^2} = 241.71 V/m[/tex]
The direction from the y-axis will be:
[tex]\beta = arctan(\frac{-121.875 V/m}{208.74 V/m}) = 30.28[/tex]° to the left of E1.
the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical.
To find the resultant electric field, you should add the horizontal and vertical components of E1 and E2 separately:
Vertical Component:
E1y = 99 V/m (vertical component of E1)
E2y = 164 V/m * sin(48°) ≈ 123.6 V/m (vertical component of E2)
Horizontal Component:
E2x = 164 V/m * cos(48°) ≈ 109.8 V/m (horizontal component of E2)
Add the vertical components:
Ey = E1y + E2y
Ey = 99 V/m + 123.6 V/m
Ey ≈ 222.6 V/m
Add the horizontal components:
Ex = E2x
Ex ≈ 109.8 V/m
Calculate the magnitude of the resultant electric field (E) using the Pythagorean theorem:
E = √(Ex² + Ey²)
E = √((109.8 V/m)² + (222.6 V/m)²)
E ≈ 247.8 V/m
So, the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical, as previously stated.
Learn more about Electric field here:
https://brainly.com/question/8971780
#SPJ12
An airplane flies with a constant speed of 1000 km/h. How long will it take to travel a distance of 1166700 meters?
Answer:
Time, t = 4200.23 seconds
Explanation:
Given that,
Speed of the airplane, v = 1000 km/h = 277.77 m/s
Distance covered, d = 1166700 m
Let t is the time taken by the airplane. The formula to find t is given by :
[tex]t=\dfrac{d}{v}[/tex]
[tex]t=\dfrac{1166700\ m}{277.77\ m/s}[/tex]
t = 4200.23 seconds
So, the airplane will take 4200.23 seconds to covered 1166700 meters. Hence, this is the required solution.
A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused a beverage cart to brake free and begin rolling at 20.0 km/h toward the tail of the plane. What is the velocity of the cart relative to the ground? (you do not have to convert these since they are all the same unit)
The velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.
To find the velocity of the cart relative to the ground when a plane is traveling north at 100.0 km/h with a 40.0 km/h crosswind blowing west and the cart is rolling at 20.0 km/h towards the tail of the plane, we can use vector addition to determine the resultant velocity.
1. First, break down the velocities into their horizontal (west-east) and vertical (north-south) components:
- Plane's velocity (north): 100.0 km/h
- Crosswind velocity (west): 40.0 km/h
- Cart's velocity (towards tail): 20.0 km/h
2. The horizontal component of the cart's velocity is the crosswind velocity (40.0 km/h), and the vertical component is its velocity towards the tail of the plane (20.0 km/h).
3. To find the resultant velocity, we can use vector addition by adding the horizontal and vertical components of the velocities separately:
Horizontal component: 40.0 km/h (west) - 0 km/h (east) = 40.0 km/h (west)
Vertical component: 100.0 km/h (north) - 20.0 km/h (south) = 80.0 km/h (north)
4. Now, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:
Resultant velocity = √(40.0^2 + 80.0^2)
Resultant velocity = √(1600 + 6400)
Resultant velocity = √8000
Resultant velocity ≈ 89.44 km/h
5. To find the direction of the resultant velocity, we can use trigonometry:
Direction = arctan(vertical component / horizontal component)
Direction = arctan(80.0 / 40.0)
Direction = arctan(2)
Direction ≈ 63.43 degrees north of west
6. Therefore, the velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.
An oil bath maintained at 50.5°C loses heat to its surroundings at the rate of 4.68 kJ/min. Its temperature is maintained by an electrically heated coil with a resistance of 60 operated from a 110 V line. A thermoregulator switches the current on and off. What fraction of the time will the current be turned on?
Answer:
The fraction of time for turn on is 0.3852
Solution:
As per the question:
Temperature at which oil bath is maintained, [tex]T_{o} = 50.5^{\circ}[/tex]
Heat loss at rate, q = 4.68 kJ/min
Resistance, R = [tex]60\Omega[/tex]
Operating Voltage, [tex]V_{o} = 110 V[/tex]
Now,
Power that the resistor releases, [tex]P_{R} = \frac{V_{o}^{2}}{R}[/tex]
[tex]P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min[/tex]
The fraction of time for the current to be turned on:
[tex]P_{R} = \frac{q}{t}[/tex]
[tex]12.148 = \frac{4.68}{t}[/tex]
t = 0.3852
A small company manufactures a certain product. The price-production relationship for this product is P = -0.7*D + 300, where P is the unit sales price of the product and D is the annual production (number of units produced). Suppose variable costs are $25 per unit produced and fixed costs are $10,287 . Find the break even point (minimum number of units that must be produced for a business to become profitable). Note: there are 2 roots in the breakeven equation, choose the smallest root.
Answer:42 units
Explanation:
Given
Price-production relationship=-0.7D+300
Total cost=Fixed cost+ variable cost
Total cost=10,287+25D
where D is the units produced
Total revenue[tex]=\left ( -0.7D+300\right )D[/tex]
Total revenue[tex]=-0.7D^2+300D[/tex]
For Break even point
Total revenue=Total cost
[tex]10,287+25D=-0.7D^2+300D[/tex]
[tex]7D^2-2750D+102870=0[/tex]
[tex]D=\frac{2750\pm \sqrt{2750^2-4\times 7\times 102870}}{2\times 7}[/tex]
[tex]D=41.869\approx 42[/tex] units
A person travels by car from Tucson to Phoenix at a constant speed of 75 km/hr. They then return from Phoenix to Tucson at a constant speed of 65 km/hr. What was their average velocity?
Answer:
[tex]v=0[/tex]
Explanation:
Knowing that the formula for average velocity is:
[tex]v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}[/tex]
Being said that, we know that the person's displacement is zero because it returns to its starting point
[tex]x_{2}=x_{1}[/tex]
That means [tex]x_{2}-x_{1}=0[/tex]
[tex]v=\frac{0}{t_{2}-t_{1}}=0[/tex]
what is degenerative accelerator?
Answer:
Degenerative accelerator:
The device which is used to study the brain and degenerative diseases like Alzheimers and Parkinson is called degenerative accelerator.
These accelerator have higher specific activity and it is comparable to reactor products.By using these accelerator many radio active nuclides can be produced those can not be produce by neutron reaction.
These generates synchrotron light that can be used for reveal the inorganic and organic structure.
Please help ASAP!!
A ball is dropped from the top of a 46.0 m -high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 22.0 m/s . The stone and ball collide part way up.
How far above the base of the cliff does this happen?
Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
A rifle with a mass of 0.9 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 750 m/s after the rifle is fired a. What is the momentum of the bullet after the rifle is fired? b. If external forces acting on the rifle can be ignored, what is the recoil velocity of the rifle?
Answer:
a ) 4.5 N.s
b) V =5 m/s
Explanation:
given,
mass of rifle(M) = 0.9 kg
mass of bullet(m) = 6 g = 0.006 kg
velocity of the bullet(v) = 750 m/s
a) momentum of bullet = m × v
= 750 × 0.006
= 4.5 N.s
b) recoil velocity
m × u + M × U = m × v + M × V
0 + 0 = 0.006 × 750 - 0.9 × V
V = [tex]\dfrac{4.5}{0.9}[/tex]
V =5 m/s
Final answer:
The momentum of the bullet after being fired is 4.5 kg*m/s. The rifle's recoil velocity, while ignoring external forces, is -5 m/s, indicating direction opposite to that of the bullet's motion.
Explanation:
The question asks about the momentum of a bullet after being fired from a rifle and the subsequent recoil velocity of the rifle. To solve this problem, we use the principle of conservation of momentum.
Part A: Bullet Momentum
The momentum of the bullet (pbullet) can be calculated using the formula p = m * v, where m is the mass and v is the velocity. For the bullet:
Mass of the bullet (mbullet): 0.006 kg
Muzzle velocity of the bullet (vbullet): 750 m/s
Therefore, the momentum of the bullet is:
pbullet = mbullet * vbullet = 0.006 kg * 750 m/s = 4.5 kg*m/s.
Part B: Rifle Recoil Velocity
By conservation of momentum, the total momentum before the bullet is fired is equal to the total momentum after. Since the rifle was at rest initially, its initial momentum is zero, and the total momentum after must also be zero. This means the momentum of the rifle (prifle) should be equal and opposite to that of the bullet:
Mass of the rifle (mrifle): 0.9 kg
Let the recoil velocity of the rifle be vrifle. The equation is:
0 = mrifle * vrifle + mbullet * vbullet
Solving for vrifle gives us:
vrifle = - (mbullet * vbullet)/mrifle = - (0.006 kg * 750 m/s) / 0.9 kg = -5 m/s.
The negative sign indicates that the rifle's velocity is in the opposite direction to the bullet's velocity, which is expected in the recoil motion.