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poizon [28]
4 years ago
11

758 j of heat are added to 0.750 kg of copper. how much does its temperature change?(unit=degrees c) PLEASE HELPPPPPPP MEEEEEE

Physics
1 answer:
DiKsa [7]4 years ago
5 0

Answer:

2.6^{\circ}C

Explanation:

When a substance is supplied with a certain amount of heat energy, the temperature of the substance increases according to the equation

Q=mC\Delta T

where

m is the mass of the substance

Q is the amount of energy supplied

C is the specific heat of the substance

\Delta T is the temperature change

In this problem:

Q = 758 J is the energy supplied

m = 0.750 kg is the mass of the sample

C=385 J/kg^{\circ}C is the specific heat of copper

Re-arranging the equation, we can find the increase in temperature:

\Delta T=\frac{Q}{mC}=\frac{758}{(0.750)(385)}=2.6^{\circ}C

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The thermal energy of what system does change
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Answer:

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True or False If the mass of the object increases, then the potential energy of the object decreases.​
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5 0
3 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
What is half-life? a. Half-life is a term that describes the time it takes for half of a particle to disintegrate c. Half-life i
Alisiya [41]
Well, half-life is the radioactivity of an identified isotope that decreases by half of the actual value.
So, your answer would be C. 
4 0
4 years ago
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