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labwork [276]
3 years ago
15

How much thermal energy is required to raise the temperature of 5 kg of water 50℃? The specific heat of water is 4168 J/kg*C.

Physics
1 answer:
sasho [114]3 years ago
3 0

Answer/solution:

Given :

Mass =5kg

T 1 =20 C,T 2  =100 ∘C

ΔT=100−20=80 ∘C

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(kgK)

Q=5×4200×80

=1680000 Joule.

=1680KJ

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A 3kg book falls from a 2m tall bookshelf what is the speed <br><br> PICTURE included
mafiozo [28]

Answer:

39.2m/s

Explanation:

The potential energy the book has right before it falls is equal to the kinetic energy in falling.

PE = KE

mgh = (1/2)mv

2gh=v

v=(2)(9.81)(2)

v=39.24m/s

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You drive at 195 miles down the 101 freeway at 65 miles per hour. How long did it take? time =
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According to my calculations, if you drive an average of 65 miles per hour, with 0 minutes of stop-time, you should reach your destination in 3 hours and 0 minutes.

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How many types of quarks are there?<br> a. 2<br> b. 4<br> c. 6<br> d. 8
Nostrana [21]
There are 6 types of quarks.

Answer: C. 6
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3 years ago
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Breathing on a cold window can cause it to fog up. which change of state causes this
Dominik [7]
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I know this because I learned it in science class in school.

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3 0
4 years ago
Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

6 0
4 years ago
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