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rjkz [21]
3 years ago
12

Someone help me pls I will give brainlist

Physics
2 answers:
stiks02 [169]3 years ago
8 0

Answer:

C and F

Explanation:

Because that’s how weather people measure tempature

(Brainliest?)

sergeinik [125]3 years ago
6 0

Answer:

the answer is point no. a °C , c. K and d. °F

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Whats 100+50+89+2000☹️
Serggg [28]

Answer:

2239

Explanation: Stack addition

3 0
2 years ago
Why do we add the masses together after that inelastic collision?
fomenos

Answer:An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.People sometimes think that objects must stick together in an inelastic collision. However, objects only stick together during a perfectly inelastic collision. Objects may also bounce off each other or explode apart, and the collision is still considered inelastic as long as kinetic energy is not conserved.

hope this helps have a nice day❤️

Explanation:

5 0
3 years ago
A charge of 80 coulombs passes through a circuit in 5 seconds. What is the current through the circuit?
UkoKoshka [18]

Answer:

I = 16amp

Explanation:

Charge coulomb ( Q ) = It

Where I =current in ampere

t = time = 5 seconds

80 = I × 5

I = 80/5

I = 16amp

The current through the circuit will be I = 16amp

4 0
3 years ago
Janet jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.8 s later. How high is th
Archy [21]
Initial height of platform is 15.87m or 16m.

8 0
2 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
2 years ago
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