All categories

3 years ago

How does kinetic energy affect the stopping distance of small vehicle compared to a large vehicle

1 answer:

Nadusha1986 [10]3 years ago

4 0

The kinetic energy for a large vehicle is different from that of a smaller vehicle, assuming that the vehicles are travelling at the same speed and stopping the same distance. This is because for a larger vehicle the kinetic energy is higher, as the mass for a larger vehicle, is more than the smaller vehicle.

You might be interested in

If calcium lost two electrons it would have the same number of electrons as

Anna11 [10]

If Calcium lost two electrons, it would have the same number of electrons as Argon which has 18 electrons.

3 0

3 years ago

ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,

Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

$d = \frac{(v_f + v_i)}{2} \times t$

here we know that

$v_f = 10 m/s$

$v_i = 0$

$t = 1.2 s$

now we will have

$d = \frac{(10 + 0)}{2}\times 1.2$

$d = 5 \times 1.2$

$d = 6.0 m$

4 0

3 years ago

PLEASE HELP SOLVE THIS PHOTO BELOW

polet [3.4K]

Answer:

Explanation:

The pattern is adding .5 to the cm every .1 in weight you just continue the table

8 0

3 years ago

Read 2 more answers

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

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

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

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago

Can someone pls help with the last one

Leviafan [203]

Answer:

ummmmmmmmmmmm ask your teacher or perant for help

Explanation:

so umm ask an adult or teen figure and maybe you will get it write not sure do

6 0

3 years ago