In order to give a spaceship at rest in a specific reference frame s a speed increment of 0.500c, seven increments are required. Then, in this new frame, it receives an additional 0.500c increment.
The speed of an object, also known as v in kinematics, is a scalar quantity that refers to the size of the change in that object's position over time or the size of the change in that object's position per unit of time. The distance travelled by an object in a certain period of time divided by the length of the period gives the object's average speed in that period.
The spacecraft moves at v1 = 0.5c after the initial increment.The equation becomes V2 = V+V1/1+V*V1/c after the second one. 2 V2 = 0.5c+0.50c/1+(0.50c)^2/c^ 2 = 0.80c
Likewise, V3 = 0.929c
V4 = 0.976c
V5 = 0.992c
V6 = 0.99c
V7 = 0.999c
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Answer:
K = -½U
Explanation:
From Newton's law of gravitation, the formula for gravitational potential energy is;
U = -GMm/R
Where,
G is gravitational constant
M and m are the two masses exerting the forces
R is the distance between the two objects
Now, in the question, we are given that kinetic energy is;
K = GMm/2R
Re-rranging, we have;
K = ½(GMm/R)
Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;
K = -½U
This question can be solved using the concept of friction energy.
The thermal energy change is b "258.4 J".
The change in thermal energy will be equal to the friction energy produced during the motion of the box.
where,
μ = coefficient of kinetic friction = 0.4
f = force applied = 38 N
d = distance traveled by the box = 17 m
Therefore,
<u>E = 258.4 J</u>
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It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.
From the question given above, the following data were obtained:
Height (H) = 206 m
<h3>Time (t) =? </h3>
NOTE: Acceleration due to gravity (g) = 10 m/s²
The time taken for the ball to get to the ground can be obtained as follow:
H = ½gt²
206 = ½ × 10 × t²
206 = 5 × t²
Divide both side by 5
Take the square root of both side
<h3>t = 6.42 s</h3>
Therefore, it will take 6.42 s for the ball to get to the ground.
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