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HACTEHA [7]
3 years ago
12

(7%) Problem 5: A thermos contains m1 = 0.73 kg of tea at T1 = 31° C. Ice (m2 = 0.095 kg, T2 = 0° C) is added to it. The heat ca

pacity of both water and tea is c = 4186 J/(kg⋅K), and the latent heat of fusion for water is Lf = 33.5 × 104 J/kg. [email protected]
Physics
1 answer:
klio [65]3 years ago
8 0
Could you explain a little more?
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State the equation of momentum impulse theorem.
Paladinen [302]

Answer:

J = Δp

Explanation:

The impulse-momentum theorem says that the impulse J is equal to the change in momentum p.

J = Δp

3 0
3 years ago
A horse is running at 12m/s accelerated to 38m/s in 10 seconds. What is the horses acceleration.
ahrayia [7]

Answer:

A horse is running at 12m/s accelerated to 38m/s in 10 seconds. What is the horses acceleration.

2.6m/s^2

3 0
3 years ago
A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse
bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, f=4.2\times 10^{14}\ Hz

Time, t=3.2\times 10^{-11}\ s

(a) Let \lambda is the wavelength of this light. It can be calculated as :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}

\lambda=7.14\times 10^{-7}\ m

or

\lambda=714\ nm

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

n=f\times t

n=4.2\times 10^{14}\times 3.2\times 10^{-11}

n = 13440

Hence, this is the required solution.

8 0
2 years ago
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin
Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

=  - GMm/ (R+h)

Potential difference

= GMm/ R -  GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed  to launch an object from the surface of Earth to a height h above the surface.

Extra  energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0
3 years ago
. Two people are pushing a car of mass 2000 kg.
Viktor [21]

Answer:

two people who are not going to be able to make it to class today because of the day and then I will be there at the house and then we can go

5 0
3 years ago
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