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schepotkina [342]
4 years ago
7

Which type of galaxy often contains the largest relative amount of gas and dust?

Physics
1 answer:
Vikki [24]4 years ago
4 0
 the answer is spiral galaxies 
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Which of the following is true statement about isotopes?
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A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4
irina [24]

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

mvL = (I + mL^2)\omega

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega

1.75 = [4.5 + 0.245]\omega

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\omega = 0.37 rad/s

so the final angular speed will be 0.37 rad/s

4 0
3 years ago
A 14.8-g piece of Styrofoam carries a net charge of -0.742 µC and is suspended in equilibrium above the center of a large, horiz
MAXImum [283]
Base on your question where a 14.8g of piece of Styrofoam carries a net charge of -0.742C and is suspended in equilibrium above the center of a large, horizontal sheet of plastic so the ask of the problem is to calculate  the charge per unit area on the plastic sheet. The answer would be 21.96
3 0
3 years ago
A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri
fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0
4 years ago
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