Ask question

Ask question

All categories

Pachacha [2.7K]

2 years ago

5

At what point on the hill will the car have zero gravitational potential energy?

2 answers:

Mila [183]2 years ago

8 0

At the bottom of the hill

zimovet [89]2 years ago

6 0

At the bottom of the hill

You might be interested in

A rolling ball has an initial velocity of 1.6 meters per second. if the ball has a constant acceleration of 0.33 meters per seco

lukranit [14]

We have:

Initial velocity (u) = 1.6 m/s
Constant acceleration (a) = 0.33 m/s²
Time (t) = 3.6 sec

There are five constant acceleration equations that would help us to find the velocity:

$v=u+at$
$s=ut+ \frac{1}{2}at^2$
$s= \frac{1}{2}(u+v)t$
$v^2=u^2+2as$
$s=vt- \frac{1}{2}at^2$

Since we have $u, a, t$ and we want $v$
We will use the first formula $v=u+at$

$v=1.6+(0.33)(3.6)$
$v= 2.788$ m/s

7 0

2 years ago

What is the speed of a jet plane that flies 7200 km in 9 hours (in km/hr)?

e-lub [12.9K]

Explanation:

800 km per her because you have to divide 7200 by 9

8 0

2 years ago

Which is the best insulator? metal, glass, plastic, or Styrofoam?

Mashutka [201]

Answer:

Styrofoam would be the best insulator because it traps the air in small pockets, blocking the flow of heat energy.

Explanation:

6 0

2 years ago

A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=35^{2}- 2 \times 9.8 \times h$

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0

2 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

2 years ago