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My name is Ann [436]

1 year ago

6

A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t

he spring exerts on the car is given by the equation F=−Kx2, where K=5000 Nm2. The potential energy stored in the spring when the car is pushed against it is most nearly:________________

1 answer:

dybincka [34]1 year ago

5 0

Answer:

1.56 J

Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

F = - k x

k = 5000 N/m

Potential energy stored = ?

energy stored in the spring

$PE = \dfrac{1}{2}kx^2$

$PE = \dfrac{1}{2}\times 5000\times 0.025^2$

PE = 1.56 J

Hence, the potential energy stored in the car is equal to 1.56 J.

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A 50-kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the

wel

Answer:

a) amount of kinetic energy converted to internal energy = 2.5 x 10 raised to power 7 Joule

b) Kinetic energy gained by the earth = 2.1 x 10-16J

c) All the kinetic energy is converted to internal energy and the energy is further converted to thermal energy hence the reason for the hotness at around where the meteorite strikes.

Explanation:

The detailed steps and appropriate application of the law of conservation of momentum is as shown in the attached file.

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2 years ago

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. controlling the angle of the sp

8090 [49]

V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m

GOOD LUCK AND HOPE IT HELPS U

6 0

1 year ago

When Royce was 10 years old, he had a mass of

coldgirl [10]

Answer:

<em>The gravitational force between Royce and Earth would be doubled at 16 years.</em>

Explanation:

<em>"Newton's law of universal gravitation states that gravitation force between two masses is proportional to the magnitude of their masses and inverse-squared of their distance".</em>

Royce Scenario

At the age of 10 Royce's mass = 30kg

At the age of 16 Royce's mass = 60kg

From Newton's law of universal gravitation, an Increase in the mass of a body would amount to a corresponding increase in the gravitational force.

In the case of Royce, the mass double between the age of 10 and 16, so there would be an increase of the gravitation force by double.

6 0

1 year ago

A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di

Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

r₁ = 0.01 m

diameter of second piston, d₂ = 0.15 m

r₂ = 0.075 m

mass of the jack to lift = ?

now,

$\dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}$

$\dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}$

$\dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}$

$F_2= \dfrac{250}{0.01^2}\times {0.075^2}$

F₂ = 14062.5 N

F = m g

$m = \dfrac{F}{g}$

$m = \dfrac{14062.5}{9.8}$

m = 1435 Kg

hence, the correct answer is option B

5 0

1 year ago

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

8 0

2 years ago