How long does it take to travel to Mercury while going 100,000 km per hour?

When the weight of the object increase block what is the force of friction applied? Explanation?

erik [133]

Answer:

There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.

8 0

2 years ago

Estimate the average rate of change in elevation from d to points i, ii, and iii, assuming the distance from d to i is 3500 m, t

Anit [1.1K]

Ummm i am not going to be able say i am high

8 0

3 years ago

A barge floating in fresh water (rho = 1000 kg/m^3) is shaped like a hollow rectangular prism with base area A = 550 m^2 and hei

patriot [66]

Answer:

Explanation:

A )

When empty , H₀ length of barge is inside water .

volume of barge inside water = A x H₀

Weight of displaced water = AH₀ x ρ x g

Buoyant force = weight of displaced water = AH₀ ρg

B)

It should balance the weight of barge

Weight = buoyant force

Weight = AH₀ ρg

mass of barge = weight / g

weight / g = AH₀ ρ

= 550 x .55 x 1000

= 302500 kg

6 0

3 years ago

6. A skier starts from rest at the top of a frictionless incline of height 20.0 m. At

n200080 [17]

Explanation:

a. KE at bottom = PE at top

½ mv² = mgh

v = √(2gh)

v = √(2 × 9.8 m/s² × 20.0 m)

v = 19.8 m/s

b. Work by friction = PE at top

mgμ d = mgh

d = h / μ

d = 20.0 m / 0.210

d = 95.2 m

6 0

3 years ago

Estimate the average force that a baseball pitcher's hand exerts on a 0.145-kg baseball as he throws a 40-m/s pitch. Assume the

Kitty [74]

Answer:

38.67 N

Explanation:

v = Final velocity = 40 m/s

u = Initial velocity

m = Mass of ball = 0.145kg

s = Displacement of ball = 3 m

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{40^2-0^2}{2\times 3}=\frac{800}{3}\ m/s^2$

F=ma

$\\\Rightarrow F=0.145\times \frac{800}{3}\\\Rightarrow F=38.67\ N$

∴ Average force that a baseball pitcher's hand exerts on the ball is 38.67 N

6 0

3 years ago

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