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kaheart [24]
3 years ago
6

Suppose that you are holding a cup of coffee in your hand identify all forces on the cup and reaction to each force

Physics
2 answers:
Sergio039 [100]3 years ago
5 0
Newton’s 3rd law should be correct:)
Stels [109]3 years ago
3 0

Answer:

Newton's third law.

Explanation:

•your hand of holding UP/DOWN the cup has a reaction with force

•Coffee inside the cup is pushing outward to the edges of the cup

•The cup is pushing inward onto the coffee

"His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A."

You might be interested in
A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0
3 years ago
A 20 cm square frame that can rotate about the 00' axis is placed in a homogeneous magnetic field with 0.5T induction directed v
IRISSAK [1]

The solution to this task is f00king Dis.

6 0
2 years ago
PLS HELP I HAVE EXAM ILL GIVE U BRAINLIST
stiv31 [10]

Explanation:

False

electron and proton attract each other

3 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
The periodic table arranges elements by increasing _________.
Olegator [25]

Answer:

Atomic number

Explanation:

Hope it helps you in your learning process.

4 0
3 years ago
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