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OleMash [197]
2 years ago
10

The specific heat of copper is 0.39 j/g*degrees celcius how much heat is needed to raise the temperature of 1000.0 g of copper f

rom 25.0 degrees celsius to 45.0 degrees celcius
Physics
1 answer:
lapo4ka [179]2 years ago
6 0
In order to calculate the amount of energy needed, we will apply:
heat energy = mass x specific heat capacity x change in temperature
Q = mcΔT

Substituting the given values,
Q = 1000 x 0.39 x (45 - 25)
Q = 7,800 Joules of heat energy are needed
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The unit of electric current is the _____.<br><br> A) volt<br> B) ohm<br> C) ampere<br> D) faraday
professor190 [17]

C

the unit of electric current is the ampere, often shortened to amp


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Which process is responsible for changing the composition of rock?
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Erosion, weathering, mechanical changes, chemical changes.

Really, any interaction can change the composition of a rock whether it be done by man or through nature.

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Describe the process you used to build a model
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Step 1 : Get your supply list together

Step 2 : Pick what model you want to do

Step 3 : Ask for a partner 

Step 4 : Complete  the model and take your time.

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3 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0
3 years ago
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