Answer: In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.
Explanation: SI unit: watt (W)
In SI base units: kg⋅m2⋅s−3
Derivations from other quantities: P = E/t; P = F...
27.9 idkkkk look it up on photomath
Answer:
<em>The answer to your question is </em><em>more force</em>
Explanation:
<em>A consequence of more mass having more inertia is that more force is required to bring the helicopter to the same speed as the bullet </em>
<u><em>I hope this helps and have a good day!</em></u>
Answer:
The final velocity of the object is 330 m/s.
Explanation:
To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:
F = ma
If we plug in the values that we are given in the problem, we get:
42 = 7 (a)
To solve for a, we simply divide both sides of the equation by 7.
42/7 = 7a/7
a = 6 m/s^2
Next, we should write out all of the information we have and what we are looking for.
a = 6 m/s^2
v1 = 0 m/s
t = 55 s
v2 = ?
We can use a kinematic equation to solve this problem. We should use:
v2 = v1 + at
If we plug in the values listed above, we should get:
v2 = 0 + (6)(55)
Next, we should solve the problem by performing the multiplication on the right side of the equation.
v2 = 330 m/s
Therefore, the final velocity reached by the object is 330 m/s.
Hope this helps!
Answer:
20.96 m/s
Explanation:
Using the equations of motion
y = uᵧt + gt²/2
Since the puck slides off horizontally,
uᵧ = vertical component of the initial velocity of the puck = 0 m/s
y = vertical height of the platform = 2 m
g = 9.8 m/s²
t = time of flight of the puck = ?
2 = (0)(t) + 9.8 t²/2
4.9t² = 2
t = 0.639 s
For the horizontal component of the motion
x = uₓt + gt²/2
x = horizontal distance covered by the puck
uₓ = horizontal component of the initial velocity = 20 m/s
g = 0 m/s² as there's no acceleration component in the x-direction
t = 0.639 s
x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m
For the final velocity, we'll calculate the horizontal and vertical components
vₓ² = uₓ² + 2gx
g = 0 m/s²
vₓ = uₓ = 20 m/s
Vertical component
vᵧ² = uᵧ² + 2gy
vᵧ² = 0 + 2×9.8×2
vᵧ = 6.26 m/s
vₓ = 20 m/s, vᵧ = 6.26 m/s
Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s
