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bixtya [17]
3 years ago
5

You and a friend are playing tug-of-war with a massless rope. You are pulling with 50 Newtons of force while your friend is pull

ing with 40 Newtons of force.
What is the tension in the rope?
What is your net force?
What is your friend’s net force?
Physics
1 answer:
const2013 [10]3 years ago
4 0

Answer:

Explanation:

A tug of war is game where two teams will be pulling a rope horizontally,

So,

Given that,

You applied a force of F1 = 50N

You friend you are pulling against applied a force of F2 = 40N

A. Tension on the rope.

Let the tension on the rope be T

The tension cause by you on the rope is the force you applied

T = 50N.

The tension cause by your friend is also

T = 40N

Adding the two tension together

T+T = 50+40

2T = 90

T = 45

OR you will see the rope as a pulley system, i.e. we will assume that the two friend force cause a tension on the rope, their are two tension on the rope, and two forces are applied

Then, tension is equal to force applied

T+T = 40+50

2T = 90

T = 45N

b. Net force by you

In this case we will take your position as the positive direction, I.e. the net force cause by you

Using newton second law

Fnet. = ΣFx

Fnet = F1—F2

Fnet = 50—40

Fnet = 10 N

c. Fnet by your friend

In this case we will take your friends position as the positive direction, I.e. the net force cause by your friend.

Fnet. = ΣFx

Fnet = F2—F1

Fnet = 40—50

Fnet = —10 N

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In the diagram, q1=+6.25 * 10^ -8 C. What is the potential difference when you go from point A to point B? Include the correct s
Nimfa-mama [501]

Answer:

Moving a unit "positive" test charge from A to B will result in a reduction in potential

V = K Q / R      potential at a point

V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q

V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8

V2 - V1 = -4.17 * 562.5 J/C

V = - 2346 Volts

7 0
3 years ago
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold
nata0808 [166]

Answer:

We apply force to move the brick.

Explanation:

Let me first of define a force .

A force is something applied to an object or thing to change it's internal or external state.

Now if a brick is resting on smooth wood inclined at 30° to the horizontal for us to overcome the friction which is also a force we have to apply a force greater than the gravity force acting on the body and then depending on the direction of the applied force the angle to apply it also.

3 0
3 years ago
What characteristics do invertebrates have in common
Volgvan
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3 0
3 years ago
A wire carries a current of 10 amps in a direction of 90 degrees with respect to the direction of an external magnetic field of
Oksi-84 [34.3K]

Answer:

15 N

Explanation:

The magnetic force on a piece of current-carrying wire is given by:

F=ILB sin \theta

where

I is the current in the wire

L is the length of the piece of wire

B is the magnetic field strength

\theta is the angle between the direction of B and I

In this problem:

I = 10 A

B = 0.3 T

L = 5 m

\theta=90^{\circ}

Substituting into the equation, we find

F=(10 A)(0.3 T)(5 m) sin 90^{\circ}=15 N

6 0
3 years ago
2. A 16.0 kg canoe moving to the left at 12.0 m/s makes an elastic head-on collison with a 4.0 kg raft moving to the right at 6.
Serhud [2]

Answer:

(a) 4.83 m/s to the left

(b) 1224 J

Explanation:

Using the law of conservation, the initial momentum equals the final momentum. Taking left as positive while right as negative then

momentum, p=mv where m is mass and v is velocity

After the collision, the canoe keep their same masses

m_1v_1+m_2v_2=m_1v_3+m_2v_4

m and v represent mass and velocity respectively, the subscripts 1 and 2 are the initial velocities and masses while subscripts 3 and 4 are final velocities

Substituting m_1 for 16 Kg, m_2 for 4 Kg, v_1 for 12 m/s, v_2 for -6 m/s since it's towards right and we already mentioned that right is negative, v_3 for 22.7 m/s then

(16 Kg\times 12 m/s)+(4 Kg\times -6 m/s)=(16 Kg\times v_4)+ (4 Kg\times 22.7 m/s)

v_4=\frac {77.2}{16}=4.825 m/s\approx 4.83 m/s hence direction is to the left

(b)

We know that kinetic energy is given by 0.5mv^{2} and by substituting the given values of mass and velocity, in this case not considering direction then we add the values of the two objects in question, and get the total before, and add the values of the two objects after and get the total after

Kinetic energy before collision=KE=0.5mv^{2}=0.5(16*12^{2}+4*6^{2})=1224 J

Kinetic energy after collision=KE=

0.5mv^{2}=0.5(16 Kg\times 4.83^{2}+ 4 Kg\times 22.7^{2})=1217.211 J

The two values of KE before and after collision are almost equal hence we conclude that energy is conserved.

5 0
3 years ago
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