Answer:
Work done, W = 10195.92 Joules
Explanation:
Given that,
Mass of the crate, m = 170 kg
Distance, d = 10.2 m
The coefficient of friction, 
Let W is the work done by the mover. It is given by in terms of coefficient of friction as :


W = 10195.92 Joules
So, the work done by the mover is 10195.92 Joules. Hence, this is the required solution.
Answer:
F = -49.1 10³ N
Explanation:
Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant
² = v₀² + 2 a x
Since the bullet is at rest, the final speed is zero
x = 11.00 cm (1 m / 100 cm) = 0.110 m
0 = v₀² + 2 a x
a = -v₀² / 2 x
a = -1320²/(2 0.110)
a = -7.92 10⁶ m / s²
With Newton's second law we find the force
F = m a
F = 6.20 10⁻³ (-7.92 10⁶)
F = -49.1 10³ N
The sign means that it is the force that the tree exerts to stop the bullet
Answer:
-929.5Joules
Explanation:
To get the work done by sam, we will calculate the kinetic energy of sam expressed as;
KE = 1/2mv²
m is the mass = 1100kg
v is the velocity = 1.3m/s
KE = 1/2(1100)(1.3)²
KE = 550(1.69)
KE = 929.5Joules
Since Sam is opposing the direction of movement, work done by him will be a negative work i.e -929.5Joules
W=F*D
83J=F*14
83/14=F
5.92N
Answer:
C. the same.
Explanation:
The drift speed of electrons in a circuit is the same all through the circuit.