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laiz [17]
3 years ago
11

An Earth satellite is in a circular orbit at an altitude of 500 km.

Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

Speed Unchanged.

Explanation:

As work is defined as a product of force over a distance. If the distance in altitude is constant = 500km, there's 0 change in distance and force, no work would be done by the gravitational force.

Since potential energy of the satellite is unchanged, unless there's additional internal energy source, the kinetic energy would remain unchanged, so would its speed.

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What is the density of iron if 5.0 cm3 has a mass of 39.5g
weqwewe [10]

Answer :\rho = 7.9\ g/cm^3

Explanation :

It is given that

Mass of iron, m = 39.5 g

Volume of iron, V = 5\ cm^3

So, density is :

density, \rho =\dfrac{mass}{volume}

\rho =\dfrac{39.5g}{5cm^3}

\rho = 7.9\ g/cm^3

7 0
3 years ago
a spiral spring of natural length 10.0cm has a scale pan hanging freely in its tower ends . When an object of mass of 20g is pla
weqwewe [10]

Answer:

Explanation:

There seems to be a typo in the problem statement.  It says the spring stretches to a shorter length after more mass is added.  Please check the problem statement.  I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm.  The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables.  If we divide the second equation by the first equation:

(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)

(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180

0.0180 (m + 0.050) = 0.02050 (m + 0.020)

0.0180 m + 0.0009 = 0.02050 m + 0.00041

0.00049 = 0.0025 m

m = 0.196

The pan has a mass of 0.196 kg, or 196 g.

4 0
3 years ago
Suppose you took a trip to the moon. Write a paragraph describing how and why your weight would change. Would your mass change t
noname [10]
Your weight would change but not your mass, the moon has less gravity so therefore you are going to be lighter :-)
5 0
2 years ago
g We saw in class that in a pendulum the string does no work. We also saw that the normal force does no work on an object slidin
Ymorist [56]

Answer:

From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

       i.e Normal \ Force(N)  = mg cos \theta

And \theta = 90 so

           N = 0

c

An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position        

d

A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work  

Explanation:

5 0
3 years ago
Read 2 more answers
if a stone is projected at an angle of 50 degrees to the horizontal with an initial velocity of 50m/s, what is the vertical comp
Vilka [71]

Answer:

38.3 m/s

Explanation:

To find vertical component of initial velocity, you'd have to use sine ratio:

\displaystyle{\sin \theta = \dfrac{u_y}{u}}

\displaystyle{u_y} is vertical component of initial velocity and \displaystyle{u} is initial velocity given which is 50 m/s.

A stone is projected at an angle of 50 degrees so \displaystyle{\theta} = 50°. Substitute in the formula:

\displaystyle{\sin 50^{\circ} = \dfrac{u_y}{50}}\\\\\displaystyle{50 \sin 50^{\circ} = u_y}\\\\\displaystyle{u_y = 38.3 \ \, \sf{m/s}}

Therefore, the vertical component of initial velocity is approximately 38.3 m/s

(The picture is also attached for visual reference!)

3 0
2 years ago
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