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3 years ago

A race car accelerates from 0 m/s to 20.0 m/s with a displacement of 35.0 m. What is the vehicle's acceleration?

Physics

1 answer:

shutvik [7]3 years ago

4 0

Explanation:

Using Equations of Motion :

${v}^{2} = {u}^{2} + 2as$

$a = \frac{{v}^{2} - {u}^{2}}{2s} = \frac{400 - 0}{70} = 5.71msec^{ - 2}$

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The international space station makes 15.65 revolutions per day in its orbit around the earth. assuming a circular orbit, how hi

sweet-ann [11.9K]

373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km

7 0

3 years ago

Read 2 more answers

1. A truck with a mass of 8, 000 kg is traveling at 26.8 m/s when it hits the brakes. A.)What is the momentum of the truck befor

NikAS [45]

Answer:

1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s

B.) The force it takes to stop the truck is approximately 17,290.4 N

Explanation:

1. A.) The given parameters are;

The mass of the truck, m = 8,000 kg

The velocity of the truck when it hits the brakes, u = 26.8 m/s

Momentum = Mass × Velocity

The moment of the truck = The mass of the truck × The velocity of the truck

Therefore;

The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s

B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck

The time it takes the truck to come to a complete stop, t = 12.4 s

The deceleration, "a" of the truck is given by the following kinematic equation of motion

v = u - a·t

Where;

v = The final velocity of the truck = 0 m/s

u = The initial velocity = 26.8 m/s

a = the deceleration of the truck

t = The time of deceleration of the truck = 12.4 s

Substituting the known values gives;

0 = 26.8 - a × 12.4

Therefore;

26.8 = a × 12.4

a = 26.8/12.4 ≈ 2.1613

The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²

Force = Mass × Acceleration

The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck

∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.

8 0

3 years ago

Alex places 2 cubes side-by-side on a ramp made of wood. Cube #1 is ice and Cube #2 is wood

zmey [24]

Answer:

Explanation:

The sandpaper block did not move because the forces of friction and gravity were balanced.

6 0

3 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

3 0

3 years ago

All objects emit ______ radiation? A-electromagnetic B-kinetic D-solar

natulia [17]

A, electromagnetic radiation

6 0

3 years ago