The international space station (ISS) orbits the earth with a velocity 7.6 km/s. How much energy is required to lift a payload o
1 answer:
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Answer:
A) The acceleration is zero
<em>B) The total distance is 112 m</em>
Explanation:
<u>Velocity vs Time Graph</u>
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.
A. To calculate the acceleration, we use the formula:
Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:
This confirms the previous conclusion.
B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.
Area of rectangle= base*height=12 s * 8 m/s = 96 m
Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m
The total distance is: 96 m + 16 m = 112 m
Answer:
2.40 x 10⁻¹³ C
Explanation:
= number of electrons = 6.25 x 10⁶
= charge on electron = - 1.6 x 10⁻¹⁹ C
= number of protons = 7.75 x 10⁶
= charge on proton = 1.6 x 10⁻¹⁹ C
Net charge is given as
Q =
+
Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)
Q = 2.40 x 10⁻¹³ C
The block is made of A) Tin, as its specific heat capacity is
Explanation:
When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by , according to the following equation
:
where is the specific heat capacity of the material.
In this problem, we have:
m = 2 kg = 2000 g is the mass of the unknown material
is the amount of energy supplied to the block
is the change in temperature of the material
Solving the equation for , we can find the specific heat capacity of the unknown sample:
And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.
Learn more about specific heat capacity:
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