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slega [8]
3 years ago
9

A 0.2-stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone

makes 150 revolutions per minute, the tension force of the string on the stone is:
Physics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

2960 N

Explanation:

Convert rev/min to rad/s:

150 rev/min × (2π rad/rev) × (1 min / 60 s) = 50π rad/s

Sum of forces in the centripetal direction:

∑F = ma

T = m v² / r

T = m ω² r

T = (0.2 kg) (50π rad/s)² (0.6 m)

T = 2960 N

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Option B is correct: It has a negative charge.

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A calorimeter directly measures ______ in order to calculate ______. (1 point)
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The complete sentence is:

A calorimeter directly measures changes in temperature in order to calculate specific heat.

In fact, the amount of energy acquired/released by a substance is directly proportional to its change in temperature due to the equation

Q=mC_s \Delta T

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2 Newtons to the right.

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7 0
3 years ago
An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

4 0
2 years ago
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