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rjkz [21]
3 years ago
6

In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of

600 newtons. What is the net force?
Physics
1 answer:
seraphim [82]3 years ago
8 0

Answer:

Explanation:

There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.

That is completely arbitrary. It could be the other way around. It does not matter.

Left is minus so: - 600 N   is the force going left.

Right plus so: + 500 N

Now just add.

Net Force = +500 - 600

Net Force = - 100 N

So the Net Force is - 100 N going to the left.

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A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet
Ede4ka [16]

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.  

7 0
3 years ago
I will GIVE BRAINLYEST AND POINTS PLEAS HELP!!!
Elenna [48]

Answer:

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Explanation:

I figured that these answers made the most sense so ya.

3 0
3 years ago
What Kinetic energy is exactly equal to Gravitational Potential Energy why is height halfway between the maximum height?
prohojiy [21]

Explanation:

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force

Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

5 0
2 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.
lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × 10^4 newtons/coulomb

We have the magnitude of the load:

q = 6.4 × 10 ^{-19} coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × 10^{-2} meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × 10^{-19})(6.5 × 10^4)(1.2 × 10^{-2})

PE = 5.0 x 10^{-16} joules

None of the options shown is correct.

6 0
3 years ago
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