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3 years ago

14

When a pitcher throws a softball to a catcher, the vibration of the atoms that make up the softball is ____________ energy, whil

e the motion of the ball toward the catcher is ___________ energy.?

1 answer:

Molodets [167]3 years ago

5 0

The vibration is thermal energy ("heat" energy which every object possesses).
The second one is kinetic energy ("motion" energy of a massive object)

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The efficiency of a carnot cycle is 1/6. If on reducing the temperature of the sink 75 degree Celsius, the efficiency becomes 1/

svp [43]

Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is $\frac{1}{6}$

So, the equation is

$\frac{1}{6} = 1 - \frac{T_2}{T_1}$

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is $\frac{1}{3}$

Therefore the next equation is

$\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}$

Now solve both the equations

solve equations (1) and (2)

$2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)$

T_2 + 450 = 75

T_2 = 375

Now put the T_2 value in any of the above equation

i.e

T_1 = T_2 + 75

T_1 = 375 + 75

= 450

7 0

4 years ago

Does an odometer in a car measure distance or displacement?

kifflom [539]

The odometer measures distance

5 0

3 years ago

Read 2 more answers

Find the equivalent resistance of this circuit

Phantasy [73]

Answer:

Req = 564 Ω

Explanation:

The equivalent resistance between R1 and R2:

1/R =1/R1 + 1/R2

1/R =1/960 + 1/640

1/R = 1/384

R = 384

Now, the equivalent resistance between R and R3:

Req = 384 + 180

Req = 564 Ω

8 0

3 years ago

Two point charges of equal magnitude (and opposite sign) are 7.5 cm apart. At the midpoint of the line connecting them, their co

Shkiper50 [21]

Comment

The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.

Equation

The field equation is, in its simplest form,

E = kq/r^2

So each of the charges are pulling / pushing in the same direction. The equation becomes.

kq/r^2 - (-kq/r^2) = Field magnitude in N/C

Givens

K = 9 * 10^9 N m^2 / c^2
E = 45 N/C
r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
Find Q

Solution

k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C Combine

2*k*q / 0.0375^2 = 45 N/C Divide by 2

kq /(0.0375^2) = 22.5 N/C Multiply by 0.0375^2

kq = 22.5 * 0.0375 ^2 Find d^2

kq = 22.5 * 0.001406 Combine

kq = 0.03164 N/C * m^2 Divide by k

q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2

q = 2.84760 * 10 ^8 C

I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.

7 0

4 years ago

8. Two positive charges (+8.0 mC and +2.0 mC) are separated by 300 m. A third charge is placed at distance r from the +8.0 mC ch

muminat

Answer:

0.20 km

Explanation:

The computation of the distance r is shown below:

This can be calculated by using the following formula

As we know that

$\frac{k_q_1}{r^2} = \frac{k_q_1}{(300 - r)^2}$

where,

q_1 is 8 mC

q_2 is 2 mC

Now placing these values

Now

$\frac{k (8 \times 10^{-3c})}{r^2} = \frac{k (2 \times 10^{-3c})}{(300 - r)^2}$

$\frac{m}{r^2} = \frac{1}{(300 - r)^2}$

$\frac{2}{r} = \frac{1}{300 - r}$

r = 600 - 2r

r = 200 m

r = 0.20 km

We simply applied the above formula to determine the distance of r

Hence, the distance r is 0.20 km

4 0

3 years ago