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Serhud [2]
3 years ago
13

A 2.5 kg object has an acceleration of 4 m/s2. What is the force of the object?

Physics
1 answer:
Bogdan [553]3 years ago
5 0

Force = Mass × Acceleration

Mass = 2.5 kg

Acceleration = 4 m/s^2

Force = 2.5 × 4

Force = 10 N

The force of the object is <u>10 N</u>

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in terms of the wavelength of the sound wave, about how far below the open end of the resonance tube is the first resonant posit
notka56 [123]

The first resonant position below the open end of the resonance tube is; <em><u>one-quarter of the wavelength</u></em>

In the event of the first resonant position in a resonance tube, there will be a maximum air displacement which is only one antinode right at the open end where the motion is constrained.

However, there will be no displacement at the closed end which means another one node right at the closed end where air is halted.

This means that the standing wave will have one-quarter of the wavelength in the test tube.

Thus;

L = ¼λ

Read more at; brainly.com/question/17086525

8 0
2 years ago
What is the relationship between the strength of an
Anastaziya [24]

Answer:

As the number of turns in the coil increases, the strength  of the electromagnet increases.

Explanation:

When current flows through a coil the coil behaves as an electromagnet. The strength of electromagnet depend the amount of current, no of turns of coil and the core of coil.

B=μ₀ N I

μ₀ = permeability of the core

N = Number of turns of the coil

I = Current flowing through the coil

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3 0
4 years ago
Use free fall in a sentance
Softa [21]
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5 0
3 years ago
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A force of 6n is exerted on a cart that is carrying a rock and produces an acceleration of 2 m/s2. if the cart has a mass of 1 k
Alex777 [14]
F=ma
m=f/a
m=6N/2 m per secon per second
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4 0
4 years ago
The electrical force on a 2-coulomb charge is 60 Newtons (N). What is the magnitude of the electric field at the place where the
kakasveta [241]

Explanation:

The given data is as follows.

            Charge (q) = 2 coulomb,        Force (F) = 60 N

Now, it is known that the relation between electric field, force and charge is as follows.

       Magnitude of electric field (E) = \frac{Force}{charge}

Hence, putting the given values into the above formula as follows.

              Electric field (E) = \frac{Force}{charge}

                                         = \frac{60 N}{2 C}

                                         = 30 N/C

Thus, we can conclude that the magnitude of the electric field at the place where the charge is located is 30 N/C.

4 0
4 years ago
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