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3 years ago

If you're going 80 mph how long does it take to 80 miles

Physics

1 answer:

natka813 [3]3 years ago

6 0

It would take you an hour, but only if you keep your speed.

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A car starts to move from rest and covers a distance of 360m in one minute. Calculate the acceleration of the car.

Romashka-Z-Leto [24]

<h2>The acceleration of car is 0.2 ms⁻²</h2>

Explanation:

When the car moves , the distance covered is calculated by the relation

S = u t + $\frac{1}{2}$ a t²

In this question u = 0 , because car was at rest initially

Thus S = $\frac{1}{2}$ a t²

here S is displacement and a is the acceleration of car

Therefore 360 = $\frac{1}{2}$ a ( 60 )²

Because time taken is one minute or 60 seconds

Therefore a = $\frac{360x2}{3600}$

or a = 0.2 m s⁻²

4 0

3 years ago

The average force applied to move an initially stationary golf cart over a period of 25 s is 500 N. What is the change in moment

Yuki888 [10]

Answer:

12500(kg*m/s)

Explanation:

F=ma=mv/t=p/t

p=F*t=500N*25 s=12500(kg*m/s)

6 0

3 years ago

1.imagine you were able to throw a ball in a frictionless environment such as outer space.once you let go of the ball,what will

Scorpion4ik [409]

Imagine you were able to throw a ball in a frictionless environment
such as outer space. Once you let go of the ball, it will travel forever
in a straight line, and at a constant speed. (At least until it bumps into
something.)

A car accelerates down the road. The reaction to the tires pushing
on the road is the road pushing on the tires.

5 0

3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$