M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).
The momentum before colision is equal with the momentum after colision:
m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem
=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
The possible magnitude for the force of static friction on the stationary cart is 72.1 N.
The given parameters:
- <em>Applied force on the cart, F = 72.1 N</em>
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Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.
F = ma
Static frictional force is the force resisting the motion of an object at rest.

where;
is the frictional force

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.
Learn more about Newton's second law of motion: brainly.com/question/25307325
Answer:
0.833
Explanation:
Power = energy / time
Power = force × distance / time
Power = force × velocity
P = (850 kg) (9.8 m/s²) (1.00 m/s)
P = 8330 W
P = 8.33 kW
The efficiency of the motor is therefore:
e = 8.33 kW / 10.0 kW
e = 0.833
Answer:
it is separated by 80 cm distance
Explanation:
As per Coulombs law we know that force between two point charges is given by

here we know that


force between two charges is given as

now we have



so it is separated by 80 cm distance
Known :
l = 7 cm
w = 4 cm
Asked :
h = ...?
Answer :
V = B triangle × h (long)
35 = ½ × 4 × h × 7
35 = ½ × 28 × h
35 = 14 h
h = 35 ÷ 14
h = 2,5 cm
Sorry if I am wrong, I only study