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abruzzese [7]
3 years ago
10

Your teacher's car can go from rest to 25 m/s (≈55 mph) in 10 seconds. The car's velocity changes at a uniform rate. Below is a

quantitative velocity vs time graph that represents the motion of the car. Determine the acceleration of the car.

Physics
1 answer:
Mice21 [21]3 years ago
5 0

If the speed of the car changes uniformly, as shown in the graph, this means that the acceleration is constant. As the acceleration is the derivative of the speed

a = \frac{\delta v}{\delta t}

Then, this means that the acceleration is the slope of the equation of the line shown in the graph.

Then, we find the acceleration using the following equation:

a = \frac{v_f- v_i}{t_2-t_1}\\\\a = \frac{25-0}{10}\\\\a = 2.5\ m/s ^ 2

Then, the acceleration of the car is 2.5\ m/s^2

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likoan [24]
The answer to this would be ocean floor
4 0
2 years ago
Calculate the energy of a photon of electromagnetic radiation whose frequency is 2.94 x 10 to the 14th xbox one
DIA [1.3K]

Answer:

1.95\cdot 10^{-19} J

Explanation:

The energy of a photon is given by:

E=hf

where

h=6.63\cdot 10^{-34}Js is the Planck constant

f=2.94\cdot 10^{14} Hz is the frequency of the photon

Substituting the numbers into the equation, we find:

E=(6.63\cdot 10^{-34} Js)(2.94\cdot 10^{14} Hz)=1.95\cdot 10^{-19} J

3 0
2 years ago
When the flashlight is in the air and the refracted ray enters the water, how does the angle of refraction compare with the angl
tigry1 [53]

Answer:

The refracted angle will be less than the angle of incidence and the speed of light is slower in the new medium (which is water) and is closer to the normal.

Try remembering this by using FST SFA (Fast Sofa)

                                             Fast -> Slow = Towards the normal

                                             Slow -> Fast = Away from normal

4 0
3 years ago
Question 9
egoroff_w [7]

Answer:

C

Explanation:

F=ma

given solution

v=12m/s a=v/t

s=6 sec =12m/s÷6sec

=2m/s^2 then we get acceleration now we will find the mass. first derive the the formula of mass by crisis cross then you will get this formula which is m=F/a

=36÷2

= 18

6 0
3 years ago
A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f
wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

5 0
3 years ago
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