What kind of medium does ocean waves have

An egg is thrown horizontally off the top row bleachers at the Brickyard,

Ludmilka [50]

Answer:

4

Explanation:

Divide 30 meters by 7.5 and you´re answer is 4. This is how I would think you solve the problem

6 0

2 years ago

A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le

dangina [55]

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>

8 0

2 years ago

Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas

Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

$I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}$

For 4 particles:

$I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

$I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}$

c)

$I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

8 0

2 years ago

to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on

Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

8 0

2 years ago

1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe

Serjik [45]

1. $I_2 = 0.14 I_1$