The car lands approximately 45.67 meters away from the base of the cliff, and its impact speed is 36.1 m/s.
Explanation:The situation described involves typing a projectile motion. Ignore any effect of air resistance for simplicity's sake. The stunt man's car exits the ramp at an angle, and continues its journey downwards under the force of gravity.
Firstly, we need to break down the initial velocity of the car into its horizontal and vertical components. Horizontal component (Ux) is governed by the formula Ux = U cos θ which results to 20 cos 20° = 18.79 m/s. Vertical component (Uy) is calculated by Uy = U sin θ which gives us 20 sin 20° = 6.84 m/s.
Following this, we need to calculate the total time the car is in the air (t), governed by the equation h = Uy * t - (1/2) * g * t², where g represents gravity (9.8 m/s²) and h is the height of the cliff (28m). Solving for t gives us approximately 2.43 seconds.
Finally, we find the horizontal distance (X) the car covers during the time it's in the air: X = Ux * t = 18.79 m/s * 2.43 s ≈ 45.67 m.
To calculate the impact speed, we take the square root of the sum of the squares of the final horizontal and vertical velocities. The final vertical velocity (Vy) = Uy + g * t = 6.84 m/s + 9.8 m/s² * 2.43 s = 30.95 m/s. The impact speed then equals √((18.79 m/s)² + (30.95 m/s)²) = 36.1 m/s.
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The vertical velocity of a projectile launch at 35 degree : O continuously decreases O continuously increases O sometimes decreases and sometimes increases O is zero O stays constant.
Answer:Sometime decreases and sometime increases.
Explanation:
Given
Projectile is launched at [tex]35 ^{\circ}[/tex]
so horizontal velocity=ucos35
vertical velocity=usin35
since gravity acts in the vertical direction, therefore, it affects only the vertical component of the projectile.
So Vertical velocity first decrease up to maximum height and then starts to increase.
Two charges q1 and q2 have a total charge of 11 C. When they are separated by 4.5 m, the force exerted by one charge on the other has a magnitude of 8 mN. Find q1 and q2 if both are positive so that they repel each other, and q1 is the smaller of the two. (For the universal constant k use the value 8.99 109 N m2/C2.)
If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.
Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.
To find the charges [tex]\( q_1 \) and \( q_2 \),[/tex] we can use Coulomb's law, which states that the magnitude of the force between two point charges is given by:
[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]
Given that the force exerted by one charge on the other has a magnitude of 8 mN (milli-Newtons) when they are separated by 4.5 m, we can set up the equation as follows:
[tex]\[ 8 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot |q_1 \cdot q_2|}{(4.5 \, \text{m})^2} \][/tex]
Solving for[tex]\( |q_1 \cdot q_2| \[/tex]), we get:
[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \, \text{N} \cdot (4.5 \, \text{m})^2}{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2} \][/tex]
[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \times 20.25}{8.99 \times 10^9} \, \text{C}^2 \][/tex]
[tex]\[ |q_1 \cdot q_2| = \frac{0.162}{8.99 \times 10^9} \, \text{C}^2 \][/tex]
[tex]\[ |q_1 \cdot q_2| = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]
Now, since both charges are positive and repel each other, we know that [tex]\( q_1 \)[/tex]is smaller than [tex]\( q_2 \)[/tex]. So, let's assume [tex]\( q_1 = x \) and \( q_2 = 11 - x \)[/tex]. Substituting these values into the equation for [tex]\( |q_1 \cdot q_2| \)[/tex], we get:
[tex]\[ x \cdot (11 - x) = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]
[tex]\[ 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]
Now, solve this quadratic equation to find x , which represents [tex]\( q_1 \).[/tex] Once we find x , we can find [tex]\( q_2 = 11 - x \).[/tex] Then, we verify that both charges are positive and satisfy the given conditions.
To solve the quadratic equation [tex]\( 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \)[/tex], we first rearrange it to the standard form:
[tex]\[ x^2 - 11x + 1.802 \times 10^{-11} = 0 \][/tex]
We can use the quadratic formula to find[tex]\( x \):[/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \), \( b = -11 \), and \( c = 1.802 \times 10^{-11} \).[/tex]
Plugging in these values:
[tex]\[ x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(1.802 \times 10^{-11})}}{2(1)} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]
[tex]\[ x \approx \frac{11 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ x \approx \frac{11 \pm 11}{2} \][/tex]
[tex]\[ x_1 \approx \frac{11 + 11}{2} \][/tex]
[tex]\[ x_1 \approx \frac{22}{2} \][/tex]
[tex]\[ x_1 \approx 11 \][/tex]
[tex]\[ x_2 \approx \frac{11 - 11}{2} \][/tex]
[tex]\[ x_2 \approx \frac{0}{2} \][/tex]
[tex]\[ x_2 \approx 0 \][/tex]
So, we have [tex]\( x_1 = 11 \) and \( x_2 = 0 \).[/tex]
Since [tex]\( q_1 \)[/tex] must be smaller than [tex]\( q_2 \),[/tex] we take [tex]\( x_1 = 11 \) as \( q_1 \) and \( x_2 = 0 \) as \( q_2 \).[/tex]
Therefore,[tex]\( q_1 = 11 \, \text{C} \) and \( q_2 = 0 \, \text{C} \)[/tex]. But we need both charges to be positive, so this solution is not valid.
Let's check the other root [tex]\( x_2 = 0 \):[/tex]
If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.
Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.
A block is placed at the top of a frictionless inclined plane with angle 30 degrees and released. The incline has a height of 15 m and the block has mass 2 kg. In the above problem the speed of the block when it reaches the bottom of the incline is 8.3 m/s 17 m/s 25 m/s 30 m/s
Answer:
The speed of the block when it reaches the bottom is 17 m/s.
(b) is correct option.
Explanation:
Given that,
Height = 15 m
Mass = 2 kg
We need to calculate the speed of the block when it reaches the bottom
Using conservation of energy
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=2gh[/tex]
[tex]v=\sqrt{2gh}[/tex]
Where, g = acceleration due to gravity
h = height
Put the value into the formula
[tex]v=\sqrt{2\times9.8\times15}[/tex]
[tex]v=17\ m/s[/tex]
Hence, The speed of the block when it reaches the bottom is 17 m/s.
For a satellite already in perfect orbit around the Earth, what would happen if: - the satellite's speed is reduced? - the satellite's mass is reduced?
Answer:
Answer
Explanation:
If the satellite is in perfect orbit around the earth then gravitational pull of the earth on the satellite and moving forward satellite is in perfect balance i.e gravitational pull does not effect the orbit as the velocity of the satellite is good enough to counter the gravitational pull of the earth and inertia makes the satellite move forward with greater velocity, but when its velocity is suddenly reduced, the gravitational pull will counter on the orbiting satellite and it will crash onto the earth.
This is same as in well of death when motorist moves in circular motion inside the well with centripetal and centrifugal forces balanced, suddenly crashes on the ground when velocity of the motorist is not enough to balance forces.
the satellite's mass has no affect on its orbit around the earth rather the mass of the earth is only affecting factor .
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g
Answer:
T₁ = 19.39N with an angle of 52° with the horizontal.
T₂ = 15.64N with an angle of 40° with the horizontal.
Explanation:
1)We calculate the weight of holiday decoration
W = m×g
W: holiday decoration weight in Newtons (N)
m: holiday decoration weight = 3 kg
g: acceleration due to gravity = 9.8 m/s²
W = m×g = 3 kg × 9.8 m/s² = 29.4 N
2)We apply Newton's first law to the system in equilibrium:
ΣFx = 0
T₁Cos52° - T₂Cos40°=0
T₁Cos52° = T₂Cos40°
T₁ = T₂(Cos40° / T₁Cos52°)
T₁ = 1,24×T₂ Equation (1)
ΣFy = 0
T₁Sin52° + T₂Sin40° - 29.4 = 0 Equation (2)
We replace T₁ of Equation (1) in the Equation (2)
1.24×T₂×Sin52° + T₂Sin40° - 29.4 = 0
1.88*T₂ = 29.4
T₂ = 29.4/1.88
T₂ = 15.64N
We replace T₂ = 15.64N in the Equation (1) to calculate T₁:
T₁ = 1.24×15.64N
T₁ = 19.39N
Bragg reflection results in a first-order maximum at 15.0°. In this case, at what angle would the second-order maximum occur?
Answer:
31.174°
Explanation:
Bragg's condition is occur when the wavelength of radiation is comparable with the atomic spacing.
So, Bragg's reflection condition for n order is,
[tex]2dsin\theta=n\lambda[/tex]
Here, n is the order of maxima, [tex]\lambda[/tex] is the wavelength of incident radiation, d is the inter planar spacing.
Now according to the question, first order maxima occur at angle of 15°.
Therefore
[tex]2dsin(15^{\circ})=\lambda\\sin(15^{\circ})=\frac{\lambda}{2d}[/tex]
Now for second order maxima, n=2.
[tex]2dsin\theta=2\times \lambda\\sin\theta=\frac{2\lambda}{2d}[/tex]
Put the values from above conditions
[tex]sin\theta=2\times sin(15^{\circ})\\sin\theta=2\times 0.258819045103\\sin\theta=0.517638090206\\\theta=sin^{-1}0.517638090206\\ \theta=31.174^{\circ}[/tex]
Therefore, the second order maxima occurs at 31.174° angle.
How would you find the total energy stored in the
capacitorsif.....
A 2.0 microF capacitor and a 4.0 microF capacitor are connected
inPARALLEL across a 300V potential difference.
Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = [tex]2.0\micro F = 2.0\times 10^{- 6} F[/tex]
C' = [tex]4.0\micro F = 4.0\times 10^{- 6} F[/tex]
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
[tex]C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F[/tex]
The energy of the capacitor, E is given by;
[tex]E = \frac{1}{2}C_{eq}V^{2}[/tex]
[tex]E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J[/tex]
Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as seen by someone outside the car? Please explain reasoning, thank you!
Answer with Explanation:
Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.
For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference
While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.
If a truck is travelling east on a straight road and travels 100 meters in 25s what is the truck's velocity?
Answer:
Velocity of the truck, v = 4 m/s
Explanation:
It is given that,
Distance covered by the truck, d = 100 m
Time taken by the truck, t = 25 s
Let v is the velocity of the truck. We know that the velocity is a vector quantity. Mathematically, it is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{100\ m}{25\ s}[/tex]
v = 4 m/s
So, the velocity of the truck is 4 m/s. Hence, this is the required solution.
A 2 gram marble is placed in the bottom of a frictionless, hemispherical bowl with a diameter of 4.9 m and it is set in circular motion around the lowest point in the bowl such that its radius of "orbit" is 16 cm. What is its speed? (If you find this problem confusing, consider a simple pendulum with a string length half the diameter given here and suppose its amplitude is 16 cm.)
Answer:
6.93 m/s
Explanation:
mass of marble, m = 2 g
Diameter of the bowl = 4.9 m
radius of bowl, r = half of diameter of the bowl = 2.45 m
The length of the string is 2.45 m
Now it is executing simple harmonic motion, thus the potential energy at the bottom is equal to the kinetic energy at the highest point.
[tex]mgr=\frac{1}{2}mv^{2}[/tex]
where, m be the mass of marble, v be the velocity of marble at heighest point, g be the acceleration due to gravity and r be the radius of the path which is equal to the length of the string
By substituting the values
9.8 x 2.45 = 0.5 x v^2
v = 6.93 m/s
Thus, the speed is 6.93 m/s.
A mechanics shop has a 16,500 N car sitting on a large car lift of 125 cm^2 piston. How much force is needed to be applied to the small piston of area 4.51 cm^2 to lift it?
Answer:595.32 N
Explanation:
Given
Mechanic shop has a car of weight([tex]F_1[/tex]) 16,500 N
Area of lift([tex]A_1[/tex])=[tex]125 cm^2[/tex]
Area of small piston([tex]A_2[/tex])=[tex]4.51 cm^2[/tex]
According to pascal's law pressure transmit will be same in all direction in a closed container
[tex]P_1=P_2[/tex]
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
[tex]F_2=F_1\frac{A_2}{A_1}[/tex]
[tex]F_2=16,500\times \frac{4.51}{125}=595.32 N[/tex]
A standard 1 kilogram weight is a cylinder 48.0 mm in height and 55.0 mm in diameter. What is the density of the material?
Answer:8769 Kg/m^3
Explanation: First of all we have to calculate the volume of the cylinder so it is equal to:
Vcylinder= π*r^2*h where r and h are the radius and heigth, respectively.
Vcylinder= π*0.0275^2*0.048=1.14*10^-4 m^3
The density is defined by ρ=mass/volume
the we have
ρ=1Kg/1.14*10^-4 m^3=8769 Kg/m^3
At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt on which you can either stand or walk. Suppose a speed ramp has a length of 121 m and is moving at a speed of 2.2 m/s relative to the ground. In addition, suppose you can cover this distance in 78 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 121 m using the speed ramp?
Answer:
It takes you 32.27 seconds to travel 121 m using the speed ramp
Explanation:
Lets explain how to solve the problem
- The speed ramp has a length of 121 m and is moving at a speed of
2.2 m/s relative to the ground
- That means the speed of the ramp is 2.2 m/s
- You can cover the same distance in 78 seconds when walking on
the ground
Lets find your speed on the ground
Speed = Distance ÷ Time
The distance is 121 meters
The time is 78 seconds
Your speed on the ground = 121 ÷ 78 = 1.55 m/s
If you walk at the same rate with respect to the speed ramp that
you walk on the ground
That means you walk with speed 1.55 m/s and the ramp moves by
speed 2.2 m/s
So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s
Now we want to find the time you will take to travel 121 meters using
the speed ramp
Time = Distance ÷ speed
Distance = 121 meters
Speed 3.75 m/s
Time = 121 ÷ 3.75 = 32.27 seconds
It takes you 32.27 seconds to travel 121 m using the speed ramp
What fraction of all the electrons in a 25 mg water
dropletmust be removed in order to obtain a net charge of 40
nC?
Answer:
[tex]9.11\times 10^{-15}.[/tex]
Explanation:
The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.
The charge on one electron, [tex]\rm e=-1.6\times 10^{-19}\ C.[/tex]
Let the N number of electrons have charge -40 nC, such that,
[tex]\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.[/tex]
Now, mass of one electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex]
Therefore, mass of N electrons = [tex]\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.[/tex]
It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.
Let it is m times the total mass of the droplet which is [tex]25\ \rm mg = 25\times 10^{-6}\ kg.[/tex]
Then,
[tex]\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.[/tex]
It is the required fraction of mass of the droplet.
Since vectors always have positive magnitudes, the sum of two vectors must have a magnitude greater than the magnitude of either one of them. O True O False
Answer:
False.
Explanation:
Yes the magnitude of a vector is always positive , but a vector consists of
magnitude directionwhen two vectors are added their direction may be opposite to each other For example-
[tex]a=3i+3j[/tex]
[tex]b=3i-6j[/tex] ,
then their resultant
[tex]r=b+a\\r=3i+3j+3i-6j\\r=6i-3j[/tex]
This resultant vector's x and y component equal to y and x component of vector b so its magnitude will be equal to magnitude of vector b.
Therefore, the resultant magnitude not necessary equal to the magnitude of either vector.
The neutron mass is 1.675 x 10-27 kg. Express the proton mass in eV/c, keeping 4 significant figures B) The mass of the electron is 9.109 x 10-31 kg. Express this in eV/c . C) What is the mass of the hydrogen atom in eV/c??
Answer:
Explanation:
given,
mass of neutron = 1.675 × 10⁻²⁷ kg
1 kg = 5.58 × 10³⁵ eV/c²
a) mass of proton = 1.675 × 10⁻²⁷ × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²
b) mass of electron = 9.109 x 10⁻³¹ × 5.58 × 10³⁵ = 5.08 × 10⁵eV/c²
c) mass of hydrogen = 1.675 × 10⁻²⁷ × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²
A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car passes the first sign at t = 1.3 s, the second sign at t = 3.9 s, and the third sign at t = 5.5 s.
(a) What is the magnitude of the average velocity of the car during the time that it is moving between the first two signs?
(b) What is the magnitude of the average velocity of the car during the time that it is moving between the second and third signs?
(c) What is the magnitude of the acceleration of the car?
The average velocity between the first two signs is 9.1 m/s, between the second and third signs is 10.4 m/s, and the car's acceleration is approximately 0.26 m/s².
To solve for the average velocity and acceleration of the car passing three equally spaced traffic signs, we can use the kinematic equations for uniformly accelerated motion.
Part a: average velocity between the first two signs
The average velocity is calculated by dividing the displacement (distance between the signs) by the time interval: vavg1,2 = d /
[tex](t_2 - t_1)[/tex]. Substituting the given values, we have vavg1,2 = 25 m / (3.9 s - 1.3 s), resulting in an average velocity of 9.1 m/s.
Part b: average velocity between the second and third signs
Similarly, for the average velocity between the second and third signs: vavg2,3 = d / ([tex]t_3 - t_2[/tex]). Using the given times, we calculate vavg2,3 = 25 m / (5.5 s - 3.9 s), giving an average velocity of 10.4 m/s.
Part c: magnitude of the acceleration
To find the acceleration, we can use the velocities we just calculated, and the time between the signs. We can set up an equation for motion for the time intervals. As initial velocity v0 for the second interval is the final velocity of the first interval, we get v0 = vavg1,2 and final velocity v = vavg2,3. Then we can use the equation v = v0 + a[tex](t_3 - t_2)[/tex] to find the acceleration. Rearranging for a, we have a = [tex](v - v_0) / (t_3 - t_2)[/tex]. After calculating, we find that the acceleration is approximately 0.26 m/s².
A 34 kg block slides with an initial speed of 9 m/s up a
rampinclined at an angle of 10o with the horizontal.
Thecoefficient of kinetic friction between the block and the ramp
is0.6. Use energy conservation to find the distance the block
slidesbefore coming to rest.
Answer:
5.4 m
Explanation:
mass, m = 34 kg
initial velocity, u = 9 m/s
final velocity, v = 0 m/s
coefficient of friction, μ = 0.6
Angle of inclination, θ = 10°
Let teh distance traveled before to come into rest is d.
According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the block.
Work done by the gravitational force = W1 = - mg Sinθ x d
Work done by the frictional force = W2 = - μ N = - μ mg Cosθ x d
negative sign shows that the direction of force and the direction of displacement is opposite to each other.
Total work done W = W1 + w2
W = - 34 x 9.8 x Sin 10 x d - 0.6 x 34 x 9.8 x cos 10 x d
W = - 254.86 d
Change in kinetic energy = 0.5 x m (v^2 - u^2)
= 0.5 x 34 (0 - 81) = - 1377
So, W = change in KE
- 254.86 d = - 1377
d = 5.4 m
An aircraft carrier is sailing at a constant speed of 15.0m/s. it is suddenly spotted by an enemy bomber aircraft flying in the same direction as the ship is moving. the bomber is flying at a constant speed of 100 m/s at an altitude of 1.00 km above the carrier when it decides to attack the carrier. if the bomber drops its bomb at a horizontal distance pf 1.3 km from the aircraft carrier, will it score a direct hit?
Answer:
Explanation:
Relative velocity = 100 - 15 = 85 m /s
Distance to be covered up with this relative speed = 1.3km
= 1300m
If t be the time taken
85 x t = 1300
t = 1300/ 85
= 15.29
Time of fall of the bomb t₁ by 1 km
1000 = 1/2 g t₁²
t₁² = 2000/9.8
t₁ = 14.28 s
Since t₁≠ t₂
It will not score a direct hit.
A proton is accelerated down a uniform electric field of 450 N/C. Calculate the acceleration of this proton.
Answer:
The acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]
Explanation:
We have given electric field E = 450 N/C
Charge on proton [tex]=1.6\times 10^{-19}C[/tex]
Force on electron due to electric field is given by [tex]F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N[/tex]
Mass of electron [tex]m=1.67\times 10^{-27}kg[/tex]
Now according to second law of motion [tex]F=ma[/tex]
So [tex]720\times10^{-19}=1.67\times 10^{-27}a[/tex]
[tex]a=431.137\times 10^8m/sec^2[/tex]
So the acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]
HOW DOES LIGHT BEHAVE MOVING FROM A MORE DENSE MEDIUM TO ALESS
DENSE MEDIUM?
Answer:
Light travels faster in a less dense medium
Explanation:
In a less dense medium, light will travel faster than it did in the more dense medium (an example of this transition can be light going from water to air).
This is because in a less dense medium there ares less particles, so light finds less resistence in its path.
The opposite happens in a more dense medium, there are more particles so light will travel slower because more particles will get in the way.
In summary, the more dense the medium, the more particles it has, and the slower that light will travel in it.
A 2.00- kg ball has zero potential and kinetic energy.
Mariadrops the ball into a 10.0- m deep well. After the ball comes
to astop in the mud, the sum of its potential and kinetic
energyis?
Answer:
- 196.2 J
Explanation:
Given:
Mass of the ball = 2.00 kg
Height through which ball drops = 10.0 m
Since it is mentioned that the ball comes to a stop in the mud, therefore the velocity of the ball will be zero i.e v = 0 m/s
Thus.
the kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
= [tex]\frac{1}{2}m\times0^2[/tex] = 0 J
Now,
The potential energy = - mgh
here, g is the acceleration due to gravity
and, the negative sign depicts that the ball is falling down
thus,
The potential energy = - 2 × 9.81 × 10 = - 196.2 J
Hence, the total energy = - 196.2 + 0 = - 196.2 J
Final answer:
The sum of the gravitational potential and kinetic energy of the 2.00-kg ball after it comes to a stop in the mud at the bottom of a 10.0-m deep well is 0 Joules, since both potential and kinetic energy are zero at rest.
Explanation:
When the 2.00-kg ball is dropped into a 10.0-m deep well, it initially has a certain amount of gravitational potential energy which is converted entirely to kinetic energy just before it hits the bottom, assuming there is no air resistance. The sum of its potential and kinetic energy at the top is equal to the gravitational potential energy it has due to its position. This can be calculated using the formula for gravitational potential energy (GPE), which is GPE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and h is the height above the reference point. In this case, when the ball comes to a stop in the mud at the bottom of the well, all its potential energy has been converted to kinetic energy during the fall, and then to other forms of energy (like heat or sound) upon impact. The ball's potential energy at the bottom of the well is zero and the kinetic energy is also zero since the ball is at rest, so the sum of its potential and kinetic energy after coming to a stop is:
mgh = 2*9.81*10 = -196.2 J (as the ball is falling down)
Sum = -196.2J
According to the conservation of energy principle, energy is neither created nor destroyed, only transformed from one form to another. Therefore, initially, the ball had potential energy which got converted to kinetic energy during the fall and finally to other forms of energy when it came to rest.
What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s
6.0 m/s
18 m/s
20 m/s
36 m/s
Answer: 20 m/s
Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:
Ekinetic=(1/2)*(m*v^2)
then E=0.5*30Kg*(20 m/s)^2=15*400=6000J
If there is no air resistance, a quarter dropped from the top of New York’s Empire State Building would reach the ground 9.6 s later. (a) What would its speed be just before it hits the ground? (b) How high is the building?
Answer:
a) 94.176 m/s
b) 452.04 m
Explanation:
t = Time taken by the quarter to reach the ground = 9.6 s
u = Initial velocity
v = Final velocity
s = Displacement
a)
[tex]v=u+at\\\Rightarrow v=0+9.81\times 9.6\\\Rightarrow v=94.176 m/s[/tex]
Speed of the quarter just before it hits the ground is 94.176 m/s
b)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 9.6^2\\\Rightarrow s=452.04\ m[/tex]
The building is 452.04 m high
A motorist drives north for 35.0 min at 85.0 km/h and then stops for 15.0 min. He then continues north, traveling 130 km in 2.00 h. (a) what is his total displacement? b) what is his average velocity?
Final answer:
The motorist's total displacement is 179.58 km north, and his average velocity is approximately 63.38 km/h north for the entire trip.
Explanation:
Displacement and Average Velocity Calculation
For the given problem, we can find the motorist's total displacement by adding the distances traveled in each segment of the journey, all in the same direction (north). First, the motorist drives north for 35.0 minutes at 85.0 km/h. The distance for this part of the trip is:
Distance = Speed × Time = 85.0 km/h × (35.0 min / 60 min/h) = 49.58 km (rounded to 2 decimal places)
The next segment involves a stop, so the displacement remains unchanged.
Finally, the motorist drives an additional 130 km north.
Therefore, the total displacement is the sum of these distances: Total Displacement = 49.58 km + 130 km = 179.58 km north.
To find the average velocity, we need to consider the total displacement and the total time, including the stop:
Total time = Driving time + Stopping time = (35.0 min + 2.00 h + 15.0 min) = 2.833 hours (2 hours and 50 minutes).
Average Velocity = Total Displacement / Total Time = 179.58 km / 2.833 h ≈ 63.38 km/h north.
The motorist's total displacement is 179.58 km north, and his average velocity over the entire trip is approximately 63.38 km/h north.
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobaric heating to a temperature of 935.9°C. (a) What is the final volume of the gas?(in cm^3 ) (b) It is then isothermally compressed to a volume 29.3cm^3; what is its final pressure?(in Pa )
Answer:
Final volumen first process [tex]V_{2} = 98,44 cm^{3}[/tex]
Final Pressure second process [tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
Explanation:
Using the Ideal Gases Law yoy have for pressure:
[tex]P_{1} = \frac{n_{1} R T_{1} }{V_{1} }[/tex]
where:
P is the pressure, in Pa
n is the nuber of moles of gas
R is the universal gas constant: 8,314 J/mol K
T is the temperature in Kelvin
V is the volumen in cubic meters
Given that the amount of material is constant in the process:
[tex]n_{1} = n_{2} = n [/tex]
In an isobaric process the pressure is constant so:
[tex]P_{1} = P_{2} [/tex]
[tex]\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }[/tex]
[tex]\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }[/tex]
[tex]V_{2} = \frac{T_{2} V_{1} }{T_{1} }[/tex]
Replacing : [tex]T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}[/tex]
[tex]V_{2} = 98,44 cm^{3}[/tex]
Replacing on the ideal gases formula the pressure at this piont is:
[tex]P_{2} = 3,92 * 10^{9} Pa[/tex]
For Temperature the ideal gases formula is:
[tex]T = \frac{P V }{n R }[/tex]
For the second process you have that [tex]T_{2} = T_{3} [/tex] So:
[tex]\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }[/tex]
[tex]P_{2} V_{2} = P_{3} V_{3} [/tex]
[tex]P_{3} = \frac{P_{2} V_{2}}{V_{3}} [/tex]
[tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
Considere un campo vectorial U que apunta radialmente hacia afuera de un punto P. La magnitud del campo en un punto Q una distancia r del punto P es a K donde K es una constante positiva. Calcule el flujo de este campo vectorial a través de una superficie esférica de radio R centrada en el punto P
Answer:
El flujo del campo U a través de dicha superficie es [tex]\Phi_U=4\pi r^2K[/tex]
Explanation:
El flujo vectorial de un campo U a través de una superficie S está dado por
[tex]\Phi_U=\int_S \vec{U} \cdot \hat{n} \ dS[/tex]
donde [tex]\hat{n}[/tex] representa el vector unitario normal a la superficie. Teniendo en cuenta que el campo se encuentra en dirección radial, se tiene
[tex]\Phi_U=\int_S K\hat{r} \cdot \hat{r} \ dS = K\int_S dS = KS=K*4\pi r^2[/tex]
A 0.676 m long section of cable carrying current to a car starter motor makes an angle of 57.7º with the Earth’s 5.29 x 10^−5T field. What is the current when the wire experiences a force of 6.53 x 10^−3 N?
Answer:
The current in the wire under the influence of the force is 216.033 A
Solution:
According to the question:
Length of the wire, l = 0.676 m
[tex]\theta = 57.7^{\circ}[/tex]
Magnetic field of the Earth, [tex]B_{E} = 5.29\times 10^{- 5} T[/tex]
Forces experienced by the wire, [tex]F_{m} = 6.53\times 10^{-3} N[/tex]
Also, we know that the force in a magnetic field is given by:
[tex]F_{m} = IB_{E}lsin\theta[/tex]
[tex]I = \frac{F_{m}}{B_{E}lsin\theta}[/tex]
[tex]I = \frac{6.53\times 10^{-3}}{5.29\times 10^{- 5}\times 0.676sin57.7^{\circ}[/tex]
I = 216.033 A
The pilot of an airplane reads the altitude 6400 m and the absolute pressure 46 kPa when flying over a city. Calculate the local atmospheric pressure in that city in kPa and in mmHg. Take the densities of air and mercury to be 0.828 kg/m3 and 13,600 kg/m3, respectively.
The local atmospheric pressure in the city in kPa is ?
The local atmospheric pressure in the city in mmHg is ?
Answer:
1. 6.672 kPa
2. 49.05 mm of mercury
Explanation:
h = 6400 m
Absolute pressure, p = 46 kPa = 46000 Pa
density of air, d = 0.823 kg/m^3
density of mercury, D = 13600 kg/m^3
(a) Absolute pressure = Atmospheric pressure + pressure due to height
46000 = Atmospheric pressure + h x d x g
Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa
(b) To convert the pressure into mercury pressure
Atmospheric pressure = H x D x g
Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity
6672 = H x 13600 x 10
H = 0.04905 m
H = 49.05 mm of mercury
To calculate the local atmospheric pressure in the city, use the formula P = P0 * e^(-Mgh/RT), where P is the pressure at the current location, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Plug in the given values into the formula and solve for P0 to find the local atmospheric pressure in kPa. To convert this pressure to mmHg, use the conversion factor that 101.325 kPa is equal to 760 mmHg.
Explanation:To calculate the local atmospheric pressure in the city, we can use the relationship between altitude, pressure, and density of air. Altitude affects atmospheric pressure because as we go higher in the atmosphere, the air becomes less dense and therefore exerts less pressure. The formula to calculate pressure is given by:
P = P0 * e^(-Mgh/RT)
Where P is the local atmospheric pressure, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Since we know the altitude and pressure at the current location, we can rearrange the formula to solve for P0:
P0 = P / e^(-Mgh/RT)
Using the given values, we can substitute them into the formula. The density of air is 0.828 kg/m³, the altitude is 6400 m, the absolute pressure is 46 kPa, the molar mass of air is the average molar mass of air (28.97 g/mol), the acceleration due to gravity is 9.8 m/s², the ideal gas constant is 8.314 J/(mol·K), and the temperature can be assumed to be the average temperature at that altitude (around 0°C or 273 K).
Plugging in these values into the formula, we get:
Calculating this expression will give us the local atmospheric pressure in kPa.
To convert this pressure to mmHg, we can use the conversion factor that 101.325 kPa is equal to 760 mmHg. So, to find the pressure in mmHg, we can multiply the pressure in kPa by the conversion factor.
The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest street legal cars, the Bugatti Veyron Super Sport, goes from 0 to 60 in 2.70 s. To the nearest integer, what is its acceleration in m/s^2? Please show work.
Answer:
9.934 m/s²
Explanation:
Given:
Initial speed of the Bugatti Veyron Super Sport = 0 mi/h
Final speed of the Bugatti Veyron Super Sport = 60 mi/h
Now,
1 mi/h = 0.44704 m / s
thus,
60 mi/h = 0.44704 × 60 = 26.8224 m/s
Time = 2.70 m/s
Now,
The acceleration (a) is given as:
[tex]a=\frac{\textup{Change in speed}}{\textup{Time}}[/tex]
thus,
[tex]a=\frac{26.8224 - 0 }{2.70}[/tex]
or
a = 9.934 m/s²