All categories

4 years ago

Please help I'm desperate. and please explain how need to show work!

Physics

1 answer:

ehidna [41]4 years ago

6 0

Answer:

33.2 m/s

Explanation:

In order to solve this problem, we need to know the mass of the car.

Here, we assume that the mass of the car is

m = 1 kg

Then, we can apply the work-energy theorem, which states that the work done on the car by the braking force is equal to the change in its kinetic energy:

$W=\Delta E_k\\Fd=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

F = -3.87 N is the force applied by the brakes (negative because its direction is opposite to the motion of the car)

d = 4.92 m is the displacement of the car

u = 33.8 m/s is the initial velocity of the car

v is the final velocity of the car

Therefore, solving for v, we find:

$v=\sqrt{\frac{2Fd}{m}+u^2}=\sqrt{\frac{2(-3.87)(4.92)}{1}+(33.8)^2}=33.2 m/s$

You might be interested in

Does kinetic energy stay the same at all heights? pls help I have 12 minutes to finish the project and the teacher won't help me

Alex Ar [27]

Answer:

Kinetic energy does not stay the same at all heights

Explanation:

Well as the height and wind increase so does the kinetic energy it's like when you fall as you are about to hit the floor you speed increases

HOPE THIS HELPS YA :)

7 0

3 years ago

A 3-ton Toyota land cruiser travelled at a speed of 45m/s, collides with a 2- ton comfort taxi which was parked in front of the

Olin [163]

Answer:

v = 27 m/s

Explanation:

To find the speed of cars after the collision you take into account the momentum conservation law. Total momentum of both cars before the collision must be equal to the total momentum of both cars after the collision.

After the collision both cars traveled together, then you have:

$m_1v_1+m_2v_2=(m_1+m_2)v$ (1)

m1: mass of the Toyota = 3-ton = 3000 kg

m2: mass of the taxi = 2-ton = 2000kg

v1: speed of the Toyota before the collision = 45m/s

v2: speed of the car before the collision = 0 m/s (it is at rest)

v: speed of both cars after the collision = ?

You solve the equation (1) for v:

$v=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Next, you replace the values of the rest of the variables:

$v=\frac{(3000kg)(45m/s)+0kgm/s}{3000kg+2000kg}=27\frac{m}{s}$

hence, just after the collision both cars have a speed of 27m/s

7 0

3 years ago

The the figure shows a famous roller coaster ride. You can ignore friction. If the roller coaster leaves Point Q from rest, what

ch4aika [34]

Answer:

22 m/s

Explanation:

PEf +KEf =PE0 +KE0 →PE0 −PEf =KEf

−mgΔy= 1 mv2 →v= −2gΔy = −2(9.8 m/s2)(−25 m)=22 m/s

3 0

3 years ago

A car accelerates uniformly from rest at 2.2 m / s^2 for 3.0 s.Calculate the speed of the car at time t = 3.0 s

Alexxx [7]

Answer:

6.6m/s

Explanation:

we know that

v = u + at

= 0+ 2.2*3

=6.6m/s

4 0

4 years ago

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

6 0

4 years ago