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just olya [345]
3 years ago
12

A man weighs 1190 n on earth. what would he weigh on jupiter, where the free-fall acceleration is 25.9 m/s 2 ? the acceleration

of gravity is 9.8 m/s 2 . answer in units of n.
Physics
1 answer:
Elden [556K]3 years ago
8 0
The answer would be 3,145N. Using W=mg solve for the mass of the man on earth. Once you have the mass you can multiply it by the gravity of Jupiter giving you his weight in Newton’s on Jupiter.
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1.Ben is pushing a block up a ramp that is 3 meters high and 6 meters long. The block has a force of 49N. It is taking Ben 37 Ne
inna [77]
Idrk sorry... i’m doing this bc it’s asking me to. wish i could help you guys . i really do !
5 0
3 years ago
Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that
lawyer [7]

As we know by the first law of thermodynamics

Q = \Delta U + W

here we know that

Q = heat given to the system

\Delta U = U_f - U_i

W = work done by the system

now here we can say

\Delta U = 180 - 78 = 102 J

W = 64 J

now we can say that heat will be given as

Q = 64 + 102 = 166 J

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/
weqwewe [10]

Answer:Given mass = 4.1kg

Radius = 0.0117m

Velocity V = 8.4m/s

Coefficient of friction = 0.25

Explanation: Below is an attached solution to the problem stated above.

1. The angular acceleration is equal to 524rad/s^2

2. The linear acceleration is equal to 2.45m/s^2

3.the time it takes the ball to begin rolling = 0.98s

4. The distance the ball slides before it begins to roll = 7.05m

3 0
3 years ago
A 2.0-μF capacitor and a 4.0-μF capacitor are connected in series across a 1.0-kV potential. The charged capacitors are then dis
vekshin1

Answer:

Explanation:

Given that,

We have two capacitors connected in series

C1=2.0-μF

C2=4.0-μF

Then the equivalent of their series connection

1/Ceq = ½ + ¼

1/Ceq= (2+1)/4

1/Ceq=¾

Taking the reciprocal

Ceq= 4/3 μF

The capacitors are connected to a battery of 1kv

V=1000Volts

We know that,

Q=CV

Where Q is charge

C is capacitance and

V is voltage

Then, Q=4/3 ×1000

Q=4000/3 -μC

Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF

After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,

For parallel connection, they have the same voltage but different charges.

When connected in parallel, there is a charge redistribution,

And the total charge will be 2•4000/3=8000/3 -μF

Then, Q1 +Q2= 8000/3 μF

Now the charge on each capacitor will be, let them have a common voltage V

Q=CV

Then, Q1=C1V

Q1= 2×V=2V

Q2= 4×V=4V

Then, Q1+Q2=8000/3

4V+2V=8000/3

6V=8000/3

V=8000/(3×6)

V=4000/9

V=444.44Volts

Now, Q1=2V

Q1=2×4000/9

Q1=8000/9 μF

Also, Q2=4V

Q2=4×4000/9

Q2=16000/9 μF

4 0
2 years ago
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