How long will it take a car to speed up from 10m/s to 34m/s if it can accelerate 8m/s/s in the direction it is moving?

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Fynjy0 [20]

D.
The reading between 7N and 8N would have to be 7.5N. Answers A and B are much to small and answer C is way to big.

8 0

3 years ago

A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio

goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

$r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}$

$r_{cm} = {(37.26, 0)}{14.8}$

$r_{cm} = (2.52 cm, 0)$

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

$r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}$

$r_g_{cm} = {(37.26, 0)}{14.8}$

$r_g_{cm} = (2.52 cm, 0)$

So center of mass is same as center of gravity because value of gravity is constant here

3 0

3 years ago

true or false? short-wave heat radiation is given off by the earth when the earth absorbs long-wave radiation.

Greeley [361]

Answer:

Hope this helps u, pls mark me brainlist

Explanation:

The sun emits shortwave radiation because it is extremely hot and has a lot of energy to give off. Once in the Earth's atmosphere, clouds and the surface absorb the solar energy. The ground heats up and re-emits energy as longwave radiation in the form of infrared rays.

4 0

2 years ago

An apple with a mass of 0.95 kilograms hangs from a tree branch 3.0 meters above the ground. If it falls to the ground, what is

lubasha [3.4K]

Ke = pe
pe = mgh
= 0.95 x 9.8 x 3
= 27.93 J

6 0

3 years ago

Read 2 more answers

A diver jumps off a cliff 50m high and needs to clear the rock that extend outward 5.0m from the base of the cliff. The diver ju

igor_vitrenko [27]

Answer:

He should run at least at 1.5 m/s

The diver will enter the water at an angle of 87° below the horizontal.

Explanation:

Hi there!

The position and velocity of the diver are given by the following vectors:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity vector at time t

Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.

We know that, to clear the rocks, the position vector r final (see figure) should be:

r final = ( > 5.0 m, -50 m)

So let´s find first at which time the y-component of the vector r final is - 50 m:

y = y0 + v0y · t + 1/2 · g · t²

-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m

Solving the quadratic equation

t = 3.4 s

Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:

x = x0 + v0x · t (x0 = 0)

5.0 m < v0x · 3.4 s

v0x > 5.0 m / 3.4 s

v0x > 1.5 m/s

The initial horizontal velocity should be greater than 1.5 m/s

To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:

vfx > 1.5 m/s

Now, let´s calculate vfy:

vfy = v0y + g · t

vfy = 2.1 m/s - 9.8 m/s² · 3.4 s

vfy = -31 m/s

Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:

$|v| = \sqrt{(1.5 m/s)^{2} + (31m/s)^{2}} = 31 m/s$

Then the final velocity of the diver will be greater or equal than 31 m/s.

To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:

cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s

θ = 87°

The diver will enter the water at an angle of 87° below the horizontal.

8 0

3 years ago