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ANEK [815]
1 year ago
10

Which statement is true of a piece of ice at 0°C that is put into a freezer at -18°C?

Physics
1 answer:
butalik [34]1 year ago
3 0

The statement which is true about a piece of ice at 0°C which is put into a freezer at -18°C is it having the temperature of the freezer.

<h3>What is Temperature?</h3>

This is referred to the degree of hotness or coldness of a body and the unit is Celsius or Kelvin.

The ice at 0°C will experience a change in temperature of the freezer when put in it in this scenario.

Read more about Ice here brainly.com/question/2267329

#SPJ1

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the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec
vivado [14]
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3 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after
Delicious77 [7]
\Delta L=  \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

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5 0
3 years ago
Read 2 more answers
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
A rising pendulum bob gains _____ energy.
puteri [66]

Answer:

B. Kinetic energy.............

4 0
3 years ago
Read 2 more answers
Determine the amount of work done when a crane lifts a 100-N block form 2m above the ground to 6m above the ground
Dima020 [189]

Data given:

F=100N

Δx=6m-2m=4m

A=?

Formula needed:

A=F×s

Solution:

A=100N×4m

A=400J

4 0
3 years ago
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