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Dmitriy789 [7]
3 years ago
8

This force involves the attraction between objects with mass.

Physics
2 answers:
yKpoI14uk [10]3 years ago
3 0
B.) Gravitational Mass
Vlad [161]3 years ago
3 0

Gravitational mass. An easy way to remember is that if something is gravitational, it ATTRACTS.

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A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

                  = 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0
3 years ago
How much work is done in holding a 20 N sack of potatoes while waiting in line at the
Yuki888 [10]

Answer: A

Explanation:

honestly, it sounded the best

8 0
2 years ago
HELP PLEASE I need to finish this asap
ELEN [110]

Answer:

I'm not 100% sure, but I think the answer would be the first one because there's a force pushing the object in every direction, so they would cancel eachother out and make the object stay in the same place.

Explanation:

pls vote brainliest

6 0
3 years ago
Read 2 more answers
Which two formulas are used to calculate potential and kinetic energy
goblinko [34]

Answer:

gravitational potential energy:

GPE = m g h

kinetic energy:

KE = 1/2 m v^2

6 0
3 years ago
Read 2 more answers
Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done
blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

\Rightarrow a=\frac{F}{m}

Hence acceleration of the block is given by:

\Rightarrow a=\frac{25}{2}=12.5m/s^2

Displacement of the object is given by:

\Rightarrow S=ut+\frac{1}{2}at^2

Substitute the values

\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2

\Rightarrow S=56.25 m

Now work done is given by:

 W=F.S

W = 25×56.25

W =1406.25 J

3 0
3 years ago
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