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Julli [10]
3 years ago
11

A 50.0 g golf ball is struck by a driver whose head has a mass of 500.0 kg. The final speed of the golf ball after leaving the t

ee is 45.0 m/s. Find the final momentum of the golf ball.
A)2.5 kg·m/s
B)5.0 kg·m/s
C)7.5 kg·m/s
D)10.0 kg·m/s
Physics
1 answer:
Savatey [412]3 years ago
8 0

<h2>A. 2.5 kg.m/s</h2>

this is my answer sure po ako n tama iyan

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You are asked to design a retroreflector using two mirrors that will reflect a laser beam by 180 degrees independent of the inci
prisoha [69]

Answer:

Consider the diagram. We are effectively being asked to prove that $\alpha=i_1$, for any value of $i_1$. Now, from trigonometry,

Explanation:

3 0
4 years ago
You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of
Rzqust [24]

Answer:

<h2>3.36J</h2>

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

                                       = 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0
3 years ago
The diagram shows how Earth is tilted on its axis as it revolves around the Sun. What is the season in the United States when Ea
Fudgin [204]

Winter

It is winter because the Northern Hemisphere is not facing the sun.

5 0
3 years ago
The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0
3 years ago
Why is there an upward force on objects in water?
Paraphin [41]
If you'r referring to some objects, it means that the mass of the object is less than the water so it floats. If the mass of an object is greater than the mass of the water, it will sink. Compare it to a balloon, helium makes it rise, while normal air makes it sink.
4 0
3 years ago
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