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myrzilka [38]
3 years ago
12

A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c

ools to 40.0 C? The specific heat capacity of copper is 0.377 j/(g.C)
Physics
1 answer:
Zolol [24]3 years ago
6 0

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

Q=mC_s \Delta T

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

\Delta T=40.0 C - 80.0 C=-40.0 C is the change in temperature of the wire

Substituting into the equation, we find

Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J

And the sign is negative because the heat is released by the wire.

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The magnification of a microscope is increased when_________.
azamat

Answer:

Option B

Explanation:

Magnification of Microscope is  

M = M_o \times M_e  

Mo= Magnification of objective lens and  

Me= magnification of the eyepiece.  

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.  

Magnification,  

M \propto \dfrac{1}{f}

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B

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3 years ago
The Back Saver Sit and Reach from the Fitnessgram measure this fitness component.
MAVERICK [17]
The correc answer is B.
7 0
3 years ago
In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m
Serga [27]

Answer:

q = 6.48 \times 10^{-14} C

Explanation:

Deflection in the drop is due to electric field force

so we will have

F = qE

acceleration of the drop is given as

a = \frac{qE}{m}

a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}

a = 7.75 \times 10^{15} q

now we know that time to cross the plates is given as

t = \frac{D}{v}

t = \frac{0.02}{18}

t = 1.11 \times 10^{-3} s

now the deflection is given as

d = \frac{1}{2}at^2

0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2

0.310 \times 10^{-3} = 4.78 \times 10^9 q

q = 6.48 \times 10^{-14} C

5 0
2 years ago
A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s
Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

5 0
3 years ago
Which items do NOT describe possible solutions to air pollution?
bonufazy [111]

Answer:

2, 6

Explanation:

2 because if you cut down more trees you will have less items to help collect co2

6 because if you have more manufracturing more gasses will be release and moe carbon dioxcide in the air so it will slowly kill the ozone layer.

6 0
3 years ago
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