Answer:
0.025 m
Explanation:
From the question,
Applying Hook's law
F = ke................... Equation 1
Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.
make e the subject of the equation
e = F/k....................... Equation 2
Given: F = 10 N, e = 395 N/m
Substitute these values into equation 2
e = 10/395
e = 0.025 m
Answer:
E=930.84 N/C
Explanation:
Given that
I = 1150 W/m²
μ = 4Π x 10⁻⁷
C = 2.999 x 10⁸ m/s
E= C B
C=speed of light
B=Magnetic filed ,E=Electric filed
Power P = I A
A=Area=4πr² ,I=Intensity




E=930.84 N/C
Therefore answer is 930.84 N/C
A string wound around a cylinder of 10 cm<span> radius has a 150 gram mass attached. When released, the mass accelerates at 50 </span>cm/s2<span>.</span>
Answer:
your neighbors biycyle inner tubes
Explanation:
hope this helps :)
The correct answer to the question is 2.27
i.e the acceleration of the body is 2.27
along the forward direction.
CALCULATION:
As per the question, the net external force on the propeller of model airplane F = 6.8 N.
The mass of the model air plane m = 3.0 kg
We are asked to calculate the acceleration of the air plane.
From Newton's second law of motion, we know that the net external force acting on a body is equal to the product of mass with acceleration of that body.
Mathematically force F = m × a
⇒ 

[ans]
The direction of acceleration is along the direction of force. Hence, the acceleration of the propeller is 2.27
along forward direction.