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ivolga24 [154]
2 years ago
7

Mickey, a daredevil mouse of mass m , m, is attempting to become the world's first "mouse cannonball." He is loaded into a sprin

g-powered gun pointing up at some angle, and is shot into the air. The gun's spring has force constant k k and is initially compressed a distance X X from its relaxed position. If Mickey has a constant horizontal speed V V while he is flying through the air, how high above his initial location in the gun does Mickey soar? Find an expression for Mickey's maximum height h m hm in terms of m , m, V , V, X , X, k , k, and g .
Physics
2 answers:
Sati [7]2 years ago
8 0

Answer:

  h = v₀² / 2g ,      h = k/4g     x²

Explanation:

In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches

Starting point. Lower compressed spring

              Em₀ = K = ½ m v²

Final point. Highest on the path

             Em_{f} = U = mg h

             

As or no friction the energy is conserved  

              Em₀ =  Em_{f}

              ½ m v₀²² = m g h

             h = v₀² / 2g

We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse

                  ½ k x² = 2 g h

                  h = k/4g     x²

quester [9]2 years ago
6 0

Answer:

 h = v₀² / 2g ,      h = k/4g     x²

Explanation:

edg

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An object can be broken up by a planet's gravity once it passes the _______. The Jovian planets are composed primarily of ______
Rina8888 [55]

Answer:1. Roche limit

2.hydrogen

3.atmosphere

4.mercury

5.venus

6.when an object passes the Roche limit, the strength of gravity on the object increases. If the density of the planet is higher, then the object can break up farther away from the planet. If the density is lower, then the Roche limit is located closer to the planet

7.Farther our in the solar system, beyond the frost line, hydrogen was at a low enough temperature that it could condense. This allowed hydrogen to accumulate under gravity, eventually forming the Jovian planets

Explanation:

3 0
2 years ago
Read 2 more answers
A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
2 years ago
A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
Mekhanik [1.2K]
The initial volume of the gas is
V_i = 600 cm^3
while its final volume is
V_f = 200 cm^3
so its variation of volume is
\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3

The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
Q=+100 J

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
\Delta U = Q-W = 100 J - (-160 J)=+260 J
where the positive sign means the internal energy of the gas has increased.
7 0
2 years ago
A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
netineya [11]

Answer:

E=\frac{\lambda}{2\pi r\epsilon_0}

Explanation:

We are given that

Linear charge density of wire=\lambda

Radius of hollow cylinder=R

Net linear charge density of cylinder=2\lambda

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

\oint E.dS=\frac{q}{\epsilon_0}

q=\lambda L

E(2\pi rL)=\frac{L\lambda}{\epsilon_0}

Where surface area of cylinder=2\pi rL

E=\frac{\lambda}{2\pi r\epsilon_0}

8 0
2 years ago
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