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2 years ago

A gun has a muzzle speed of 150 m/s. Find two angles of elevation that can be used to hit a target 800m away.

Physics

1 answer:

lorasvet [3.4K]2 years ago

4 0

Answer:

Explanation:

u = 150 m/s

R = 800 m

The formula for the horizontal range is

$R=\frac{u^{2}Sin2\theta }{g}$

where, θ is angle of projection.

$800=\frac{150^{2}Sin2\theta }{9.8}$

Sin2θ = 0.348

2θ = 20.4°

θ = 10.2°

The range is same for the complementary angles.

So, the other angle is 90 - 10.2 = 79.8°

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A force does 30000 J of work along a distance of 9.5m. Find the applied force.

Elina [12.6K]

Answer: F=3158N

Explanation:

Work is the product of force applied and the distance the object moves along the force applied. Work is measured in joules and the equation is as follows

W = F x d

So that F = W/d

F = 30000 J / 9.5m

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2 years ago

A force of 6600 N is exerted on a piston that has an area of 0.010 m2

sveticcg [70]

Answer:

Choice A: approximately $0.015\; \rm m^2$ , assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

$\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}$ .

By Pascal's Principle, because the first piston exerted a pressure of $6.6\times 10^{5}\; \rm N \cdot m^{-2}$ on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be $6.6\times 10^{5}\; \rm N \cdot m^{-2}$ . In other words:

$P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}$ .

To achieve a force of $9.900 \times 10^3\; \rm N$ , the surface area of the second piston should be:

$\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}$ .

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3 years ago

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

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As per the question, the mass of meteorite [ m]= 50 kg

The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

$Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2$

$=\frac{1}{2}50kg*[1000\ m/s]^2$

$=\frac{1}{2}50* 10^{6}\ J$

$=25*10^6\ J$

Here, J stands for Joule which is the S.I unit of energy.

$Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J$

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3 years ago

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3 years ago

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