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Pavel [41]
4 years ago
8

Which kind of force do you exert on an object when you put it towrd you?

Physics
1 answer:
shutvik [7]4 years ago
3 0
Applied force, I believe!
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What is the speed of a wave, if the wavelength is 100 m and the period is 20 s?
deff fn [24]
D. 5 m/s
This is because the formula to find the speed of a wave is the wavelength divided by the period (wavelength/period). Since the wavelength is 100 m and the period is 20 s, the speed is 5 m/s.
If you are confused on the units, then think about it this way: If you put this problem in fraction form, it would be: 100 m/20 s. 100 divided by 20 is 5, but m and s are still there. Remember that this "/" is also a division symbol. Therefore, the unit will be m/s
5 0
4 years ago
Pls help im begging you
Lostsunrise [7]

Answer:

I think it's TRUE because forces change an object's motion but dont quote me on it ok? Cause I'm not a 100 percent sure

7 0
3 years ago
Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk.
stellarik [79]

Answer: The correct option is Option b.

Explanation:

Power is defined as the rate of work done by an object.

Mathematically,

P=\frac{W}{t}    .....(1)

And work done is the product of force exerted on the object times the displacement covered by that object.

Mathematically,

W=F.s

Putting this value in above equation, we get:

P=\frac{F.s}{t}

where,

P = power = ?W

F = Force exerted = 10N

s = Displacement = 400cm = 4m   (Conversion factor: 1m = 100 cm)

t = Time taken = 8s

Putting values in above equation, we get

P=\frac{10\times 4}{8}\\\\P=5W

Hence, the correct option is Option b.

7 0
4 years ago
Read 2 more answers
A loop circuit has a resistance of R1 and a current of 1.9 A. The current is reduced to 1.5 A when an additional 3.1 Ω resistor
Deffense [45]

Answer:

11.625 Ohm

Explanation:

Let V be the Voltage charge of the loop, as this is constant we know that before the resistor addition the current I is:

V/R1 = 1.9 or V = 1.9R1

After the resistor addition to series R = R1 + 3.1

I = V/R = V/(R1 + 3.1) = 1.5

We can substitute V = 1.9R1

1.9R1 = 1.5R1 + 1.5*3.1

0.4R1 = 4.65

R1 = 4.65/0.4 = 11.625 Ohm

4 0
4 years ago
You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height o
irina [24]

Answer: 20.765 m/s

Explanation:

This problem can be solved by the conservation of energy principle, this means the initial energy E_{o} must be equal to the final energy  E_{f}:

E_{o}=E_{f} (1)

Where each energy is the sum of kinetic energy K and potential energy U:

K_{o}+U_{o}=K_{f}+U_{f} (2)

Where:

K_{o}=\frac{1}{2}mV_{o}^{2}

Being m your mass and V_{o}=0 m/s your initial velocity, since the roller coaster sterted from rest.

U_{o}=mgh_{o}

Being  g=9.8 m/s^{2} the acceleration due gravity and  h_{o}=25 m your initial height

K_{f}=\frac{1}{2}mV_{f}^{2}

Being V_{f} your final velocity

U_{f}=mgh_{f}

Being h_{f}=3 m your final height

Rewritting (2):

\frac{1}{2}mV_{o}^{2}+mgh_{o}=\frac{1}{2}mV_{f}^{2}+mgh_{f} (3)

mgh_{o}=m(\frac{1}{2}V_{f}^{2}+gh_{f}) (4)

Isolating V_{f}:

V_{f}=\sqrt{2g(h_{o}-h_{f})} (5)

V_{f}=\sqrt{2(9.8 m/s^{2})(25 m-3 m)} (6)

Finally:

V_{f}=20.765 m/s This is your spedd when you arrive at 3 m height

7 0
4 years ago
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