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Mazyrski [523]
3 years ago
7

Fill in this pretty little blank.

Physics
2 answers:
IgorLugansk [536]3 years ago
6 0
Freaky girl , it from my
Trava [24]3 years ago
3 0

Answer:

freaky girl      

it from my

Explanation:

lol

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40 POINTSS. Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop u
seraphim [82]

Answer: If there is a higher friction, the opposition force is higher so that it can reduce our speed. So, a factor that affects friction is the roughness or smoothness of the surface of the object. In comparison of the table with the fabric, the fabric will have a more opposition force. As the surface of the fabric is usually rougher than the surface of a smooth table. As there is more friction on a fabric, we will feel more opposition force on our finger tip.

8 0
2 years ago
A person is pushing a box.The net external force on the 60-kg box is stated to be 90 N. If the force of friction opposing the mo
Trava [24]

Answer:

a = 1 m/s²

Explanation:

given,

mass of the person = 60 Kg

Net External force exerted = 90 N

force of friction opposing the motion = 30 N

acceleration of the box = ?

Net force = External force applied - frictional force

Net force = 90 - 30

net force = 60 N

we know

F = mass x acceleration

60 = 60 x a

a = 1 m/s²

acceleration of the box is equal to a = 1 m/s²

7 0
3 years ago
PLS HELP MEEEE (NO LINKS PLEASE)
-Dominant- [34]
I think it could be D please tell me if I’m wrong I hope you have a wonderful day ❤️
5 0
3 years ago
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Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!
krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

L = 10 Log\frac{I}{I_0}

20 = 10 Log{I}{10^{-12}}

I_1 = 10^{-10}W/m^2

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

L = 10 Log\frac{I}{I_0}

50 = 10Log\frac{I}{10^{-12}}

I_2 = 10^{-7}W/m^2

now we know that

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}

r_2 = 47 cm

so now the distance from friend must be 47 cm

8 0
3 years ago
Does the voltage of a battery affect the strength of an electromagnet?
swat32

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

7 0
3 years ago
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