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Ghella [55]
3 years ago
6

How much heat is released to freeze 47.30 grams of copper at its freezing point of 1,085°C? The latent heat of fusion of copper

is 205.0 J/g.
Physics
2 answers:
katovenus [111]3 years ago
3 0

Answer:

=9696.5J

Explanation:

The latent heat of fusion of a unit mass of copper is the heat released when it changes from its molten state to solid state of the heat absorbed to melt it.

The total energy released is calculated using MLf where M is the mass of the copper and Lf is the latent heat of fusion.

=mLf

=47.30 grams×205.0J/g

=9696.5J

Rudik [331]3 years ago
3 0

Answer:

-9697

Explanation:

thats the answer

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rusak2 [61]

Answer:

9.6

Explanation:

to convert km to miles multiply by 1.609

7 0
3 years ago
Which runner has greater kinetic energy: a 45 kg runner moving at a speed of 7 m per second or a 93 kg runner moving at a
bagirrra123 [75]

Kinetic Energy = (1/2) (mass) (speed)

First runner:  KE = (1/2) (45kg) (49 m/s)  =  1,102.5 Joules  

Second runner:  KE = (1/2) (93kg) (9 m/s)  =  418.5 Joules

The <em>first runner </em><em>has 163</em>% more kinetic energy than the second runner has.

7 0
2 years ago
The greater the of an object the more force is needed to cause acceleration
d1i1m1o1n [39]

the greater the <u>mass</u> of an object the more force is needed to cause acceleration



8 0
3 years ago
A particle moves along the x axis so that its velocity at time t is given by v(t)=
vodomira [7]

Answer:

a = 0.7267 ,  acceleration is positive therefore the speed is increasing  

Explanation:

The definition of acceleration is

         a = dv / dt

they give us the function of speed

         v = - (t-1) sin (t² / 2)

         a = - sin (t²/2) -  (t-1) cos (t²/2)  2t / 2

         a = - sin (t²/2) - t (t-1)  cos (t²/2)

the acceleration for t = 4 s

          a = - sin (4²/2) - 4 (4-1) cos (4²/2)

          a = -sin 8 - 12 cos 8

remember that the angles are in radians

          a = 0.7267

the problem does not indicate the units, but to be correct they must be m/s²

We see that the acceleration is positive therefore the speed is increasing

6 0
3 years ago
Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,
klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble=2\hat{i}-3\hat{k} N

Position vector=r=0.5\hat{j}-2\hat{k} m

a.We have to find the resulting torque on the pebble about origin.

Torque=r\times F

Substitute the values then we get

Torque= (0.5j-2k)\times (2i-3k)

Torque=-k-1.5i-4jN-m

By using i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0

b.r=2i-3k

r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k

Torque about point (2,0,-3)

\tau=(-2i+0.5j+k)\times (2i-3k)

\tau=-6j-k-1.5i+2j=-1.5i-4j-kN-m

6 0
3 years ago
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