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3 years ago

What is the danger to spacecraft and astronauts from micrometeoroids?

Physics

1 answer:

Sloan [31]3 years ago

7 0

Answer:

Suppose the micrometeoroid weighed 1 g = .001 kg

Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)

Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground

v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s

KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules

One can see that 28000 Joules could be damaging amount of energy

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One situation indicates a physical change rather than a chemical change. Which situation indicates a physical change?

lisabon 2012 [21]

The physical change would be A. Solid copper melting to form liquid copper. The rest are all examples of chemical change.

6 0

4 years ago

Read 2 more answers

If a power lifter raises a 1000N weight a distance of 2 meters in 0.5 seconds, what is his power out put?

Irina18 [472]

Answer:

P = 4000 [W]

Explanation:

In order to solve this problem, we must first determine the work, which is defined as the product of force by distance.

W = F*d

where:

W = work [J] (units in Joules)

F = force = 1000[N]

d = distance = 2 [m]

W = 1000*2

W = 2000 [J]

And power is defined as the relationship between work and the time in which the work is done.

P = W/t

P = power [W] (units of watts)

t = time = 0.5 [s]

P = 2000/0.5

P = 4000 [W]

8 0

3 years ago

A substance is round in one container and has a volume of one liter. In a much larger container, the substance is rectangular bu

insens350 [35]

Answer:

Liquid

Explanation:

Solids do not change shape, so we know it's not a solid.

Gases take the shape of their container, but they also expand to the volume of the container. The substance in this problem has the same volume in different size containers, so we know it's not a gas.

The substance must be a liquid. It takes the shape of its container but has a constant volume.

8 0

3 years ago

A ball with a mass of 0.585 kg is initially at rest. It is struck by a second ball having a mass of 0.420 kg , initially moving

Y_Kistochka [10]

Answer:

a) $(v_1)=0.3989m/s$

b) $\theta_1=80.5 \textdegree$

c) $K.E=0.036J$

Explanation:

From the question we are told that:

Initial speed of 1st ball $u_{1}=0 m/s$

Mass of 1st ball $m_1=0.585kg$

Mass of 2nd ball $m_2=0.420kg$

Initial speed of 2nd ball $u_{2}=0.270 m/s$

Final speed of 2nd ball $v_{2}=0.220 m/s$

Angle of collision $\angle=36.9 \textdegree$

Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1^2+m_2v_2^2$

The final velocity $v_1$ is given as

$0+(0.420)(0.270)=(0.585)(v_1)^2+(0.420)(0.220)^2$

$(v_1)^2=\frac{(0.420)(0.270)-(0.420)(0.220)^2}{0.585}$

$(v_1)^2=0.1591$

$(v_1)=0.3989m/s$

Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1cos\theta_1+m_2\theta_2$

$0+(0.420)(0.270)=(0.585)(1.511)cos\theta_1+(0.420)(0.220)cos36 \textdegree$

$cos\theta_1= \frac{(0.420)(0.270)-(0.420)(0.220)cos36 \textdegree}{(0.585)(0.3989)}$

$cos\theta_1=0.1656$

$\theta_1=80.5 \textdegree$

Generally the equation for kinetic energy is mathematically given by

$K.E=\frac{1}{2} mv^2$

1st Ball

$K.E=\frac{1}{2} (0.585)(0.3989)^2$

$K.E=0.0465J$

2nd ball

$K.E=\frac{1}{2} (0.420)(0.220)^2$

$K.E=0.101J$

Therefore the change in the total kinetic energy of the two balls as a result of the collision is

$0.101-0.0465$

$K.E=0.036J$

3 0

3 years ago

If the length of a pendulum is doubled,what will be the change in its time period??

Phantasy [73]

Explanation:

see the attachment.. hope it will help you

6 0

3 years ago