Answer:
electric flux is 280 Nm²/C
so correct option is D 280 Nm²/C
Explanation:
radius r = 0.50 m
angle = 30 degree
field strength = 713 N/C
to find out
the electric flux through the surface
solution
we find here electric flux by given formula that is
electric flux = field strength × area× cos∅ .......1
here area = πr² = π(0.50)²
put here all value in equation 1
electric flux = field strength × area× cos∅
electric flux = 713 × π(0.50)² × cos60
we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk
so we use cos60
electric flux = 280 Nm²/C
so correct option is D 280 Nm²/C
The Electric Potential Difference Created by Point Charges 13. Two point charges, +3.40 pC and -6.10 uC, are separated by 1.20 m. What is the electric potential midway between them?
The electric potential midway between two point charges is determined by calculating the potential due to each charge separately and adding them together. Coulomb's constant and the distances to the midpoint are used in this calculation.
The student is asking for the electric potential midway between two point charges. The charges mentioned are +3.40 pC and -6.10 uC, with a separation of 1.20 m. To calculate the potential at the midway point, the contributions of both charges to the potential have to be added algebraically since electric potential is a scalar quantity.
The electric potential due to a single point charge at a distance r is given by the formula V = k * q / r, where V is the electric potential, k is Coulomb's constant (
approximately 8.99 x 109 N*m2/C²), q is the charge, and r is the distance from the charge to the point of interest. Because the point is midway, r will be 0.60 m for both charges.
Calculating the potential for each charge separately, we add the potentials resulting from each charge to find the total electric potential at the midpoint.
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.
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.
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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.
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.
An object is thrown vertically upward and has a speed of 32.6 m/s when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.
Answer:
The maximum height is 162.67 m.
Explanation:
Suppose the total height is h.
And at the height of 2/3h the speed of an object is,
[tex]u=32.6m/s[/tex]
And the remaining height will be,
[tex]h'=h-\frac{2}{3}h\\ h'=\frac{1}{3}h[/tex]
So, according to question the initial speed is,
[tex]u=32.6m/s[/tex]
Acceleration in the upward direction is negative,
[tex]a=-9.8m/s^{2}[/tex]
And the final speed will be v m/s which is 0 m/s.
Now according to third equation of motion.
[tex]v^{2} =u^{2} -2as[/tex]
Here, v is the final velocity, u is the initial velocity, a is the acceleration, s is the displacement.
[tex]0^{2} =32.6^{2} +2(-9.8)\dfrac{h}{3} \\h=\dfrac{3\times 1062.76}{2\times 9.8}\\h=162.67 m[/tex]
Therefore, the maximum height is 162.67 m.
An airplane undergoes the following displacements: First, it flies 40 km in a direction 30° east of north. Next, it flies 56 km due south. Finally, it flies 100 km 30° north of west. Using analytical methods, determine how far the airplane ends up from its starting point.
Answer:
Distance from start point is 72.5km
Explanation:
The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:
[tex]x1=d1*cos(60), y1=d1*sin(60)\\\\ x2=x1 , y2=y1-d2\\\\ x3=x2-d3*cos(30) , y3=y2+d3*sin(30)[/tex]
Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:
[tex]d=\sqrt{x3^{2} +y3^{2} }[/tex]
Replacing the given values in the equations, the distance is calculated.
Final Answer:
The airplane ends up approximately 72.53 km from its starting point.
Explanation:
To determine how far the airplane ends up from its starting point after these displacements, we can use vector addition to find the resultant displacement. Since the movements are given in terms of directions relative to north, we can use a coordinate system where north corresponds to the positive y-axis, and east corresponds to the positive x-axis.
Let's start with the first displacement:
1. The airplane flies 40 km in a direction 30° east of north.
We can resolve this displacement into x and y components:
- The x-component (eastward) is 40 km * sin(30°) because the angle is measured from the north (y-axis).
- The y-component (northward) is 40 km * cos(30°) because the angle is with respect to the vertical (north direction).
Using the fact that sin(30°) = 1/2 and cos(30°) = √3/2:
- x1 = 40 km * 1/2 = 20 km
- y1 = 40 km * √3/2 ≈ 40 km * 0.866 = 34.64 km
Now for the second displacement:
2. The airplane flies 56 km due south.
This movement is along the negative y-axis.
- x2 = 0 km (no movement east or west)
- y2 = -56 km (southward)
For the third displacement:
3. The airplane flies 100 km 30° north of west.
- The x-component (westward) will be -100 km * cos(30°) because we are measuring the angle from the north and going west is negative in our coordinate system.
- The y-component (northward) will be 100 km * sin(30°).
Using the trigonometric values found earlier:
- x3 = -100 km * √3/2 ≈ -100 km * 0.866 = -86.6 km
- y3 = 100 km * 1/2 = 50 km
Having found the components for each displacement, we can now sum them up to find the total displacement.
Total x-component (x_total) = x1 + x2 + x3 = 20 km + 0 km - 86.6 km = -66.6 km (westward)
Total y-component (y_total) = y1 + y2 + y3 = 34.64 km - 56 km + 50 km = 28.64 km (northward)
Now, we can determine the magnitude of the resultant displacement vector using the Pythagorean theorem:
R = √(x_total^2 + y_total^2)
R = √((-66.6 km)^2 + (28.64 km)^2)
R = √(4440.96 km^2 + 820.5696 km^2)
R = √(5261.5296 km^2)
R ≈ 72.53 km
So, the airplane ends up approximately 72.53 km from its starting point.
The Jurassic Park ride at Universal Studios theme park drops 25.6 m straight down essentially from rest. Find the time for the drop and the velocity at the bottom.
Answer:
V=22.4m/s;T=2.29s
Explanation:
We will use two formulas in order to solve this problem. To determine the velocity at the bottom we can use potential and kinetic energy to solve for the velocity and use the uniformly accelerated displacement formula:
[tex]mgh=\frac{1}{2}mv^{2}\\\\X= V_{0}t-\frac{gt^{2}}{2}[/tex]
Solving for velocity using equation 1:
[tex]mgh=\frac{1}{2}mv^{2} \\\\gh=\frac{v^{2}}{2}\\\\\sqrt{2gh}=v\\\\v=\sqrt{2*9.8\frac{m}{s^2}*25.6m}=22.4\frac{m}{s}[/tex]
Solving for time in equation 2:
[tex]-25.6m = 0\frac{m}{s}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\-51.2m=-9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{\frac{51.2m}{9.8\frac{m}{s^{2}}}}=2.29s[/tex]
A ball thrown by a pitcher on a women’s softball team is timed at 56.9 mph. The distance from the pitching rubber to home plate is 47.9 ft. In major league baseball the corresponding distance is 60.5 ft. If the batter in the softball game and the batter in the baseball game are to have equal times to react to the pitch, with what speed must the baseball be thrown? Assume the ball travels with a constant velocity. [Hint: There is no need to convert units; set up a ratio.]
Answer:[tex]v_b=71.86 mph[/tex]
Explanation:
Given
Velocity of soft ball is 56.9 mph
Distance between Pitching rubber to home plate is 47.9 ft
In major league distance is 60.5 ft
Let velocity of baseball is [tex]v_b[/tex]
Let t be the time for ball to reach to batter and its reaction time
since t is same for both case
[tex]\frac{47.9}{56.9}=\frac{60.5}{v_b}[/tex]
[tex]v_b=56.9\times \frac{60.5}{47.9}[/tex]
[tex]v_b=71.86 mph[/tex]
To determine the speed at which the baseball must be thrown to allow for equal reaction times between the softball and baseball batters, we can set up a ratio using the distances from the pitching rubber to home plate in both sports and solve for the desired speed.
Explanation:
To determine the speed at which the baseball must be thrown, we can set up a ratio using the distances from the pitching rubber to home plate in softball and baseball. Since the times for the batters to react should be equal, the distance ratio is equal to the speed ratio. Therefore, we can write the proportion:
(56.9 mph)/(47.9 ft) = x/(60.5 ft)
Where x represents the speed of the baseball. To solve for x, we can cross-multiply and solve for x:
x = (56.9 mph * 60.5 ft) / 47.9 ft
Calculating the right-hand side of the equation gives us the speed of the baseball:
x ≈ 71.92 mph
Therefore, the baseball must be thrown at approximately 71.92 mph to allow for equal reaction times between the softball and baseball batters.
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The cable supporting a 1950 kg elevator has a maximum strength
of21920 N. What maximum upwardacceleration can it give the elevator
without breaking?
Answer:
1.43 m/s^2
Explanation:
Each time you see mass and force, you will probably be going to need to use Newton's second Law. This law basically shows the relationship between the force being applied on an object and its mass and acceleration:
[tex]F = m*a[/tex]
Now, the force that the cable exerts on the elevator, not only has to accelarate it, but it also has to counter gravity. The maximum tension of the cable minus the weigth of the elevator would give us the net force being applied on the elevator:
[tex]T_{cable} - W_{elevator} = m_{elevator}*a[/tex]
[tex]21920 N - 1950kg*9.81 m/s^2 = 1950 kg*a\\a = \frac{21920 N - 1950kg*9.81 m/s^2}{1950kg} = 1.43 m/s^2[/tex]
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
If the world population grows at a constant rate of 1.8% per annum, how many years will it take to double? A) 17.7 years
B) 23.4 years
C) 35.0 years
D) 38.8 years
E) 69.7 years
Answer:
after 38.8 years it will double
correct option is D 38.8 years
Explanation:
given data
population grows rate = 1.8%
to find out
how many years will it take to double
solution
we consider here initial population is x
so after 1 year population will be = (100% + 1.8% ) x = 1.018 x
and after n year population will be = [tex]1.018^{n} x[/tex]
so it will double
2x = [tex]1.018^{n} x[/tex]
take log both side
log 2 = n log (1.018)
n = [tex]\frac{log2}{log1.018}[/tex]
n = 38.853
so after 38.8 years it will double
correct option is D 38.8 years
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.
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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.
On October 21, 2001, Ian Ashpole of the United Kingdom achieved a record altitude of 3.35 km (11 000 ft) powered by 600 toy balloons filled with helium. Each filled balloon had a radius of about 0.54 m and an estimated mass of 0.27 kg. (a) Estimate the total buoyant force on the 600 balloons.
To estimate the total buoyant force on the 600 balloons filled with helium, we use Archimedes' principle. By calculating the volume of each balloon and using the formula for buoyant force, we can determine the force exerted by each balloon. Multiplying this force by the number of balloons gives us the total buoyant force.
Explanation:To estimate the total buoyant force on the 600 balloons, we need to calculate the buoyant force on each balloon and then multiply it by the number of balloons. The buoyant force on a balloon can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. In this case, the fluid is the surrounding air.
The volume of each balloon can be calculated using the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius of the balloon. In this case, r = 0.54 m. Using this formula, we can calculate the volume of each balloon to be approximately 0.653 m^3.
The buoyant force on each balloon can be calculated using the formula F = ρVg, where F is the buoyant force, ρ is the density of the fluid, V is the volume of the fluid displaced, and g is the acceleration due to gravity. Since the density of air is approximately 1.225 kg/m^3 and g is approximately 9.8 m/s^2, we can calculate the buoyant force on each balloon to be approximately 7.93 N.
Finally, to calculate the total buoyant force on the 600 balloons, we can multiply the buoyant force on each balloon by the number of balloons: F_total = F_per_balloon * number_of_balloons. Plugging in the values, we get F_total ≈ 7.93 N * 600 = 4,758 N.
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The total buoyant force on the 600 balloons that Ian Ashpole used can be calculated using Archimedes' Principle and the principle of Buoyancy. The resulting force is the difference between the weight of the air displaced by the balloons and the weight of the balloons themselves, multiplied by the number of balloons.
Explanation:This is a classic problem of Archimedes' Principle and Buoyancy, principles in Physics. In simple terms, the buoyant force on a body submerged in a fluid is equal to the weight of the fluid displaced by the body. In the case of a balloon, the buoyant force can be calculated as the difference between the weight of the air displaced by the balloon and the weight of the balloon itself. For a single balloon, this would be:
FB = (weight of the air displaced) - (weight of balloon). But we have 600 balloons, so, we multiply this force by 600 to get the total buoyant force on all the balloons. Given Ian Ashpole used 600 balloons, each with an estimated mass of 0.27 kg and a radius of about 0.54 m, we can calculate the total buoyant force on these balloons.
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Consider a physical pendulum with length of 81.9 cm and mass of 165 g. If the pendulum was released from an angle less than 10°, then calculate the period of the pendulum. (g = 9.80 m/s^2)
Answer:
The period of the pendulum is 1.816 sec.
Explanation:
Given that,
Length = 81.9 cm
Mass = 165 g
Angle = 10°
We need to calculate the period of the pendulum
Using formula of period
[tex]T = 2\pi\sqrt{\dfrac{l}{g}}[/tex]
Where, l = length
g = acceleration due to gravity
Put the value into the formula
[tex]T =2\pi\sqrt{\dfrac{81.9\times10^{-2}}{9.80}}[/tex]
[tex]T=1.816\ sec[/tex]
Hence, The period of the pendulum is 1.816 sec.
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
At the end of a race a runner decelerates from a velocity of 8.90 m/s at a rate of 1.70 m/s2. (a) How far in meters does she travel in the next 6.30 s? (Assume the deceleration of 1.70 m/s2 is constant over the full 6.30 s.)
Answer:
x=22.33m
Explanation:
Kinematics equation for constant deceleration:
[tex]x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m[/tex]
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
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.
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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 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]
which of the following best describes the objectives section of a lesson plan
A. The specific learning behaviors you are expecting from the lesson
B. How will you simplify the lesson in the event that the content is too advanced for some learners?
C. How you will go about determining the childrens strengths and weaknesses once the lesson is taught.
D. The specific steps to be taken in carrying out the lesson plan.
Answer:
A. The specific learning behaviors you are expecting from the lesson
Explanation:
Lesson plan is systematic way of approaching subject learning in schools and colleges. A Lesson plan has various section among them there is section of Objectives.
Objectives are defined precise and focused goals of the learning from the particular topic that the student must learn. It is a goal oriented method where aim is already known before its accomplishment at the end of the chapter. Hence option A seem most appropriate answer.
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.
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]
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 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
Wes stands on the roof of a building, leans over the edge, and drops a rock. Lindsay waits 1.20 s after Wes releases his rock and throws her own rock straight down at 21.0 m/s. Both rocks hit the ground simultaneously. 1) Calculate the common height from which the rocks were released. Ignore the effects of air resistance. (Express your answer to three significant figures.)
To calculate the common height from which the rocks were released, use the equations of motion. Substitute the given values and solve for the height using the equations h = (1/2)gt^2 and h = v0t + (1/2)gt^2.
Explanation:To calculate the common height from which the rocks were released, we need to use the equations of motion. Let's assume the common height is h. For Wes, the time taken to reach the ground is given as 1.20 s. Using the equation h = (1/2)gt^2, where g is the acceleration due to gravity, we can substitute the values and solve for h. For Lindsay, the time taken to reach the ground is the same, 1.20 s. Using the equation h = v0t + (1/2)gt^2, where v0 is the initial velocity, we can substitute the values and solve for h. By calculating the common height from these two equations, we can determine the height from which the rocks were released.
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]
A stone is dropped into a river from a bridge 44.0 m above the water. Another stone is thrown vertically down 1.72 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
Answer:
u₀ = 17.14 m/s
Explanation:
given,
bridge height = 44 m
initial speed of the first stone = 0 m/s
initial speed of the second stone = ?
difference after which the second stone is thrown = 1.72 s
for stone 1
[tex]h = ut + \dfrac{1}{2}gt^2[/tex]
[tex]h =\dfrac{1}{2}gt_1^2[/tex]
for stone 2
[tex]h = u_0 (t_1-t) + \dfrac{1}{2}g (t_1-t) ^2[/tex]
[tex]t_1 =\sqrt{\dfrac{1}{2}gh}[/tex]
[tex]t_1 = \sqrt{\dfrac{1}{2}\times 9.81\times 44}[/tex]
t₁ = 14.69 s
[tex]44 = u_0 \times 1.72 + \dfrac{1}{2}g\times 1.72 ^2[/tex]
u₀ = 17.14 m/s
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