Find the average acceleration of a northbound subway train that slows from 12 m/s to 9.6 m/s in 0.9 s

A stone is dropped from a bridge abd it hits the water 2.2 seconds later how high is the bridge above the water

Bess [88]

Answer:

h = 23.716 m

Explanation:

Given that,

The time taken by the stone to hit the water is, t = 2.2 s

Height of the bridge above the ground, h = ?

The distance that the body will fall through the time is given by the formula

S = 1/2 gt² m

Where,

g - acceleration due to gravity

Substituting the values in the above equation

S = 1/2 x 9.8 m/s² x (2.2 s)²

= 23.716 m

Therefore, the height of the bridge from the surface of the water is h = 23.716 m

8 0

3 years ago

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

$mv_{1i}+Mv_{2i}=(m+M)v$ (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

$v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}$

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

3 0

3 years ago

In a classical carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast en

g100num [7]

Answer:

- the speed of a person "stuck" to the wall is 14.8 m/s

- the normal force of the wall on a rider of m=54kg is 1851 N

- the minimum coefficient of friction needed between the wall and the person is 0.29

Explanation:

Given information:

the radius of the cylindrical room, R = 6.4 m

the room spin with frequency, ω = 22.1 rev/minutes = 22.1 $\frac{2\pi }{60}$ = 2.31 rad/s

mass of rider, m = 54 kg

the speed of a person "stuck" to the wall

v = ω R

= 2.31 x 6.4

= 14.8 m/s

the normal force of the wall on a rider

F = m a

a = ω^2 R

= $\frac{v^{2} }{R^{2} }$ R

= $\frac{v^{2} }{R}$

F = $\frac{mv^{2} }{R}$

= $\frac{(54)(14.8)^{2} }{6.4}$

= 1851 N

the minimum coefficient of friction needed between the wall and the person

F(friction) = μ N

W = μ N

m g = μ $\frac{mv^{2} }{R}$

g = μ $\frac{v^{2} }{R}$

μ = $\frac{gR}{v^{2} }$

= $\frac{(9.8) (6.4)}{14.8^{2} }$

= 0.29

5 0

3 years ago

A 65 N boy sits on a sled weighing 52 N on a horizontal surface. The coefficient of friction between the sled and the snow is 0.

pogonyaev

Answer:

1.40 N

Explanation:

The magnitude of the frictional force is given by:

$F=\mu N$

where

$\mu$ is the coefficient of friction

N is the magnitude of the normal reaction

The coefficient of friction for this problem is $\mu=0.012$ . The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so