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lara [203]
2 years ago
8

Hii so I am out of town unable to do this for my class can you guys use things in your own home? I will BRAINLIST!!1

Physics
2 answers:
ExtremeBDS [4]2 years ago
7 0
1.)t.v
2.) tablet
3.)nightstand
4.) led lights
5.) soft pillow
That’s all I can think of because I don’t have a lot
Luda [366]2 years ago
6 0
1. bed
2. tv
3. phone
4. chair
5. desk
6. couch
7. toilet
8. bathtub
9. lamp
10. night stand
11. blinds
12. curtains
13. candles
14. pens
15. blankets
16. pillows
17. earbuds
18. soap
19. fan
20. air conditioning
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1. A car with a mass of 2500 kg accelerates when the traffic light turns green. If the net force
soldier1979 [14.2K]

;Net force = mass of the body × acceleration of the body due to the net force

; 5000 = 2500 a...then divide both sides by 2500

; acceleration(a) = 2 m/s^2

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3 years ago
Knowing the constant g what will the gravitational force between two masses be if the gravitational force between them is 36n an
charle [14.2K]
The gravitational force between two masses is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the two masses
r is the separation between the two masses

We see that the force is proportional to the inverse of the square of the distance: F \sim  \frac{1}{r^2}
therefore, if the distance is tripled:
r'=3r
The force decreases by a factor 1/9:
F \sim  \frac{1}{(3r)^2}= \frac{1}{9}  \frac{1}{r^2}

Since the original force was 36 N, the new force will be
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3 years ago
What are two ways in which friction can be useful?
pishuonlain [190]
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6 0
3 years ago
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why are we all cheating if we payed attention in what they trying to tell us we wouldn't have to cheat but we are not paying att
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3 years ago
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A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate. The plates have a
butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance C=260\ pF

Charge q=0.155\ \mu\ C

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

V= \dfrac{Q}{C}

Where, Q = charge

C = capacitance

Put the value into the formula

V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}

V=596.2\ volts

Hence,The potential difference between the plates is 596.2 volts.

7 0
3 years ago
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