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Ostrovityanka [42]

3 years ago

12

As the mass of an object in uniform circular motion increases, what happens to the centripetal force required to keep it moving

at the same speed?

1 answer:

Dmitriy789 [7]3 years ago

7 0

Answer:

It increases, because the centripetal force is directly proportional to the mass of the rotating body.

You might be interested in

What is the change in internal energy if 60 J of heat is released from a system and 30 J of work is done on the system? Use U =

k0ka [10]

The change in internal energy of the system is +30 J

Explanation:

We can solve this problem by using the first law of thermodynamics, which states that the change in internal energy of a system is given by the equation:

$\Delta U = Q -W$

where

$\Delta U$ is the change in internal energy

Q is the heat absorbed by the system (positive if it is absorbed, negative if it is released)

W is the work done by the system (positive if it is done by the system, negative if it is done by the surroundings on the system)

Therefore, in this problem, we have

$Q=-60 J$ (heat released by the system)

$W=-30 J$ (work done on the system)

Therefore, the change in internal energy is

$\Delta U = -60 - (-30) = +30 J$

Learn more about thermodynamics:

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#LearnwithBrainly

8 0

3 years ago

Given that on Earth, gravity causes an acceleration of 9.8 m/s2, what is an acceleration of 7 g?

zzz [600]

Answer:

68.6 m/s^2

Explanation:

1 g = 9.8 m/s^2

so

7 g × 9.8m/s^2 = 68.6

7 0

3 years ago

How much time would it take for the sound of thunder to travel 1500 meters if sound travels at a speed of 330 m/s

Elis [28]

Data given:

Δx=1500m

v=330m/s

t=?

Formula:

V=Δx/t

Solution:

t=1500m/330m/s

t=4.5s

7 0

3 years ago

A Rankine cycle with one closed feedwater heater with its drain cascaded backward has a water mass flow rate through the steam g

mamaluj [8]

Answer:Draw a T-s diagram for the ideal Rankine Cycle

Explanation:

6 0

3 years ago

A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

3 years ago