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2 years ago

A free-falling golf ball strikes the ground and exerts a force on it. Which sentences are true about this situation?

Physics

2 answers:

prohojiy [21]2 years ago

6 0

Hey friends

The answer would be B

~katie

bixtya [17]2 years ago

5 0

Answer:

A. A golf ball striking the ground is a collision.

B. The ground exerts an equal force on the golf ball.

Explanation:

A collision is the mutual interaction between two or more bodies, of which at least one is in motion, with an exchange of momentum and energy.

According to Newton's third law, if the golf ball exerts a force on the ground, then the ground must exert a force of equal magnitude in the opposite direction on the golf ball.

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14. The average speed of a car was 60 m/s by the time it reached the finish line. The car moved in a straight line and traveled

Leni [432]

7.5m. Have a good day :)

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2 years ago

Question 6 Multiple Choice Worth 4 points)

allsm [11]

is iron and aluminium is there

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2 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

2 years ago

Block on inclined plane experience a force due to gravity of 300N straight down. If the slope is inclined at 67.8°to the horizon

Tems11 [23]

Answer:

The component of the force due to gravity perpendicular and parallel to the slope is 113.4 N and 277.8 N respectively.

Explanation:

Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.

Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
Fperpendicular = F*cosα = 300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N

<u><em>The component of the force due to gravity perpendicular and parallel to the slope is 113.4 N and 277.8 N respectively.</em></u>

6 0

2 years ago

A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coas

kupik [55]

Answer:

a) perfectly inelastic, b) collision is inelastic, c) elastic

Explanation:

In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved

Initial instant. Before the skater touches the frisbee

p₀ = M v₁ + m v₂

where M and m are the masses of the skater and frisbee, respectively

for the final moment they give us several possibilities, in all case the moment is conserved

p₀ = $p_{f}$

case a)

Final instant. grabs the frisbee and holds it

p_{f} = (M + m) v '

p₀ = p_{f}

We can see that this shock is perfectly inelastic, it holds the fressbee

case b)

final instant.

This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.

case c)

final instant. Grab the fressbee and resend it

$p_{f} = M v_{1f} + m v_{2f}$

this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.

5 0

2 years ago