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sertanlavr [38]
3 years ago
14

Do frictional forces act in the same direction or in the opposite direction to the applied force?

Physics
1 answer:
Evgesh-ka [11]3 years ago
3 0

Frictional forces act in the direction opposite to the MOTION. That direction could be the same OR opposite to applied force.

-- If you push a loaded heavy wagon from behind, trying to get it going faster, friction is acting against you, opposite to your force.

-- If you push a loaded rolling heavy wagon from in front, trying to make it slow down, friction is acting with you, in the same direction as your force.

-- Opposite to the motion both times.

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In Part C, how would the effects differ if you connected the galvanometer to the primary coil and the battery/switch to the seco
Katen [24]

N₂ / N₁ = 13.3

A transformer is a device that changes the voltage in the primary, causing a voltage to be induced in the secondary. The expression for this voltage ratio is

ΔV₂ = N₂ /N₁ ΔV₁

where N is the number of windings on each side and V2 and V1 are the voltages in the secondary and primary, respectively.

They state that the primary voltage in this instance is 9.0 V and the secondary voltage is 120 V.

ΔV₂ /ΔV₁ = N₂ / N₁

N₂ / N₁ = 120/9

N₂ / N₁ = 13.3

Learn more about  voltage here brainly.com/question/16774947

#SPJ4.

A 9.0-V battery (with nonzero resistance) and switch are connected in series across the primary coil of a transformer. The secondary coil is connected to a light bulb that operates on 120 V. Determine the ratio of the secondary to primary turns needed for the bell's transformer. Determine the ratio of the secondary to primary turns needed for the bells transformer. Ns/Np=?

4 0
2 years ago
a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the
aleksklad [387]
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

             H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
from the ground is

             H  =  20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

                         <span>50 - (1/2 G Q²)  =  20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add  </span><span>(1/2 G Q²)  to each side.
Then it says
                                             50  =  20Q

Divide each side by 20:          2.5  =  Q .

And there we are.  The stones pass each other

                                       2.5 seconds

after they are simultaneously launched.
</span>
5 0
3 years ago
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules (with the poten
Brilliant_brown [7]

Given that,

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules. It is moving upward with a kinetic energy of 50 Joules.

We need to find the maximum height h reached by the ball. Let at a height of 10 meters, it has a potential energy of 50 Joules. So,

mgH=50\\\\mg=\dfrac{50}{h}\\\\mg=\dfrac{50}{10}\\\\mg=5\ N ........(1)

Let at a height of h m, it reaches to a maximum height. at this point, it has a total of 100 J of energy. So,

mgh=50+50\\\\mgh=100\\\\h=\dfrac{100}{5}\\\\h=20\ m

So, the correct option is (E) "h = 20 m".

6 0
3 years ago
Why so much space in the solar system?
polet [3.4K]

Answer:

To avoid collision, you know.

4 0
2 years ago
Read 2 more answers
you leave your house and travel for 6 minutes at an average velocity of 5 m/s. How far have you traveled?​
Anvisha [2.4K]

Answer:

1800 m

Explanation:

This question requires you to calculate the distance traveled after 6 minutes at a speed of 5m/s

Given speed as 5m/s and time is 6 minutes then;

Change minutes to seconds by multiplying the minutes by 60 seconds

1 minute=60 seconds

6 minutes= 6*60 = 360 seconds

Apply the formula for speed where;

S=D/T where S is speed, D is distance and T is time

To get D , rewrite formula as S*T

where S= 5m/s and T=360s

D= 5*360 =1800m

3 0
4 years ago
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