Ask question

Ask question

All categories

3 years ago

7

A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc

ulate a) The acceleration of the box (assuming it is constant while sliding)

1 answer:

Hoochie [10]3 years ago

6 0

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v= 0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

$a=- \dfrac{20^2}{2\times 30} \ m/s^2$

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be - 6.667 m/s² .

You might be interested in

A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused

Nonamiya [84]

Answer:

The velocity $\vec{v}_{c/g}$ of the cart with respect to the ground is

$\vec{v}_{c/g}=-40\hat{x}+80\hat{y}\, km/h$

if we consider North the positive y-direction and East the positive x-direction.

Explanation:

We have for relative motion the following expression:

$\vec{v}_{c/g}=\vec{v}_{c/p}+\vec{v}_{p/g}$

Where $\vec{v}_{c/g}$ is the velocity of the cart with respect to the ground, $\vec{v}_{c/p}$ is the velocity of the cart with respect to the plane and $\vec{v}_{p/g}$ is the velocity of the plane with respect to the ground.

We find that:

$\vec{v}_{c/p}=-20\hat{y}$

$\vec{v}_{p/g}=-40\hat{x}+100\hat{y}$

Thus:

$\vec{v}_{c/g}=-20\hat{y}-40\hat{x}+100\hat{y}=-40\hat{x}+80\hat{y} \, km/h$

5 0

3 years ago

During an episode of turbulence in an airplane you feel 210 n heavier than usual.if your mass is 72 kg, what are the magnitude a

lana66690 [7]

According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.

Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.

8 0

3 years ago

Question 10 of 10

Alinara [238K]

it allows only a reduced number of electrons to flow through it.

4 0

3 years ago

Read 2 more answers

James gently releases a ball at the top of a slope But does not push the ball. Space the ball rolls down the slope. Which force

NARA [144]

On an incline, the force causing the ball to move downwards would be gravity. Additionally, the component of gravity causing this ball to move downwards would be mgsintheta.

Hope this helps!

7 0

3 years ago

Read 2 more answers

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

3 years ago

Read 2 more answers