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alisha [4.7K]
3 years ago
13

A calculator has a resistance of 22 Ω. What is the power rating for this calculator when connected to a 1.5 V battery?

Physics
1 answer:
GREYUIT [131]3 years ago
3 0

Answer:

33

Explanation:

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What is the force required accelerate a 5.0 kg object at 3.0m/s^ 2?
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The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N
6 0
3 years ago
1. A car’s velocity increases from 4.0 m/s to 36 m/s over a 4.0 second time interval. What is its average acceleration?
Finger [1]

Explanation:

36-4/4= 9 m/squared. meter per squared.

acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)

7 0
2 years ago
A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr
aleksklad [387]

Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

5 0
2 years ago
Which the answer It's science
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The answer is A. ive done a 5-k race, so its for sure 3 miles. 
3 0
3 years ago
Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with
Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is 2.728\times 10^{-3}\,\frac{m}{s^{2}} (\frac{2.784}{10000}\cdot g, where g = 9.8\,\frac{m}{s^{2}}).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (g), in meters per square second, is directly proportional to the mass of the Earth (M), in kilograms, and inversely proportional to the distance from the center of the Earth (r), in meters:

g = \frac{G\cdot M}{r^{2}} (1)

Where:

G - Gravitational constant, in cubic meters per kilogram-square second.

M - Mass of the Earth, in kilograms.

r - Distance from the center of the Earth, in meters.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972\times 10^{24}\,kg and r = 382.26\times 10^{6}\,m, then the free-fall acceleration at the orbit of the Moon is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}

g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}

6 0
2 years ago
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