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3 years ago

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can somebody help me answer this question? A car went from 110 m/s to 80 m/s in 20 seconds. What was the acceleration of the car

?

1 answer:

gogolik [260]3 years ago

4 0

Answer:

-1.5m/s²

Explanation:

Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.

So for this question you have:

V_i = 110m/s
V_f = 80m/s
t_i = 0s
t_f = 20s

which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²

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A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

aliina [53]

Answer:

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

$u_y = u sin \theta$

where

u = 21.5 m/s is the initial speed

$\theta=33.5^{\circ}$ is the angle

Substituting,

$u_y = (21.5) sin 33.5^{\circ} =11.9 m/s$

The vertical position of the ball at time t is given by

$y = h + u_y t + \frac{1}{2}gt^2$

where

h = 13.5 m is the initial heigth

$g = -9.8 m/s^2$ is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

$0 = h + u_yt +\frac{1}{2}gt^2\\0 = 13.5 +11.9 t - 4.9t^2$

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

$u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s$

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

$d= u_x t = (17.9)(3.27)=58.5 m$

3 0

3 years ago

Hello i need help with this!!

mr Goodwill [35]

Answer:

Yes

Explanation:

If lamp A burnt out there would still be a wire above it that connects lamp B and C to the power source

3 0

2 years ago

A speaker vibrates at a frequency of 200 hz what is its period

Norma-Jean [14]

Period = (1) / (frequency)

Period = (1) / (200 per second) = 0.005 second = 5 milliseconds

5 0

3 years ago

Read 2 more answers

Which best describes electrical energy? a. energy released by a chemical reaction b. energy produced by flow of electric charge

inessss [21]

<span> energy produced by flow of electric charge describes a electrical energy
because movement of electric charge do effect the work on system
so correct option is B
hope it helps</span>

4 0

3 years ago

Read 2 more answers

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