It’s gonna have to b since it’s decreasing
Answer:

the answer is A.
Explanation:
Using the laws of newton:
∑F = ma
where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.
Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:
∑F = m(0)
∑F = 0
It means that:
F -
= 0
where F is the force applied and
is the friction force. Replacing the value of F, we get:
310N -
= 0
Finally, solving for
:

Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.