This 80 kg car is moving at 20m/sec at the top where the hills radius is 100m. What is the centrifugal force?

True or false. Students with a Learners License may not receive a motorcycle endorsement.

viktelen [127]

This is a true statement

3 0

3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

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

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

6 0

3 years ago

Which has more kinetic energy, a 4.0 kg bowling ball moving at 1.0 m/s or a 1.0 kg

jolli1 [7]

Answer:

The kinetic energy of bocce ball is more.

Explanation:

Given that,

Mass of a bowling ball, m₁ = 4 kg

Speed of the bowling ball, v₁ = 1 m/s

Mass of bocce ball, m₂ = 1 kg

Speed of bocce ball, v₂ = 4 m/s

We need to say which has more kinetic energy.

The kinetic energy of an object is given by :

$E=\dfrac{1}{2}mv^2$

Kinetic energy of the bowling ball,

$E_1=\dfrac{1}{2}m_1v_1^2\\\\E_1=\dfrac{1}{2}\times 4\times (1)^2\\\\E_1=2\ J$

The kinetic energy of the bocce ball,

$E_2=\dfrac{1}{2}m_2v_2^2\\\\E_2=\dfrac{1}{2}\times 1\times (4)^2\\\\E_2=8\ J$

So, the kinetic energy of bocce ball is more than that of bowling ball.

5 0

3 years ago

A construction crane, like the one shown, has a power output of 1,500 watts. If it takes 200 seconds to lift the roof to the top

Likurg_2 [28]

Answer:7.5j

Explanation:

P=w/s

4 0

3 years ago

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

3 years ago