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3 years ago

A car accelerates from 10m/s to 20m/s over a distance of 80m. What is its acceleration?

Physics

1 answer:

Lina20 [59]3 years ago

5 0

Answer:

0.125m/s^2

Explanation:

20-10=10

10 divided by 80=0.125m/s^2

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A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago

When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea

crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = $\frac{(172\ \mu C)}{V}$

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = $\frac{(172+220)\ \mu C)}{V}$

C' = $\frac{(392)\ \mu C)}{V}$

now the dielectric constant (k) is given as:

$k=\frac{C'}{C}$

substituting the values, we get

$k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}$

k = 2.279

6 0

3 years ago

Sal sprinted 40 m to the right in 5.5s what is his average velocity

PolarNik [594]

Answer: 7.27 m/s

Explanation:

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2 years ago

The yoga term asanas are which of the following?

frutty [35]

Answer:

i guss its A

Explanation:

3 0

2 years ago

two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire

valentinak56 [21]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>

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2 years ago