Answer:
Explanation:
According to the principle of energy conservation
all kinetic energy will change into thermal energy to increase its temperature
Next, divide both sides by the object's mass
the resulting temperature change is expressed as
a 62.0 kg curler runs into a stationary 78.1 kg curler and they hold on to each other. Together they move away at 1.29 m/s, west
1 answer:
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Answer:
The duration, t = 3.53 seconds
Explanation:
The following information is provided:
The equation to calculate the time t is expressed as:
...... (1)
Where
s denotes the distance in feet
We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet
Substituting the value of s into equation (1) yields:
t = 3.53 seconds
Thus, the time taken by the object is 3.53 seconds, which provides the required answer.
The question lacks details. Here is the full question.
The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.
Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?
Answer: v = 6.5 m/s
Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:
Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:
Δx = 7(5.3)
Δx = 37.1 m
The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:
1.3 frames = 1 s
7 frames = Δt
Δt = 5.4 s
<pconsequently the="" car="" was="" driving="" at:="">v = 6.87 m/s
<pthe car="" moved="" at="" an="" estimated="">velocity of 6.87 m/s.
Consider the diagram below.
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)