Please solve it ,,,,,,,,,,................................

While traveling along a highway a driver slows from 34 m/s to 17 m/s in 6 seconds. What is the automobiles acceleration?

xxTIMURxx [149]

Answer:

-2.83 m/s²

Explanation:

Initial velocity (u) = 34 m/s
Final velocity (v) = 17 m/s
Time taken (t) = 6 seconds

❖ Acceleration is defined as the rate of change in velocity with time.

→ a = (v - u)/t

v denotes final velocity
a denotes acceleration
u denotes initial velocity
t denotes time

→ a = (17 - 34)/6 m/s²

→ a = -17/6 m/s²

<h3>→ Acceleration = -2.83 m/s²</h3>

(Minus sign implies that the velocity is decreasing.)

5 0

3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

What is the formula of finding displacement

xenn [34]

Displacement is the final position of the object minus the initial position of the object.
Xf - Xi. Displacement is not the distance of the object. If you go to the right 10m and to the left another 10m, your displacement is 0m. But your distance is 20m

8 0

4 years ago

Pablo preto lift a barbell. During which stage of the lift does Pablo do work on the barbell

Arlecino [84]

While he is lifting the dumbell as the definition of work done = moving a mass through a distance = F x d

6 0

3 years ago

Read 2 more answers

Paco pulls a 67 kg crate with 738 N and of force across a frictionless floor 9.0 M how much work does he do in moving the crate

olga_2 [115]

Answer:

W = 6642 J

Explanation:

Given that,

Mass of a crate, m = 67 kg

Force with which the crate is pulled, F = 738 N

It is moved 9 m across a frictionless floor

We need to find the work done in moving the crate. Let the work done is W. It is given by :

W = F d

W = 738 N × 9 m

= 6642 J

So, the work done is 6642 J.

5 0

3 years ago