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GarryVolchara [31]
3 years ago
7

If the mass of an object is 10 Kg and it experiences a Net force of 60 Newtons. What is the acceleration?

Physics
1 answer:
Solnce55 [7]3 years ago
3 0

Answer:

6 m/s^2

Explanation:

F = ma

60N = (10 kg)a

a = 6m/s^2

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I think it is Perpendicular

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In which atmospheric layer are almost all water-based clouds formed?
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In troposphere layer all water based cloud formed. Troposphere is the lowest layer of the atmosphere. The troposphere is slightly unstable layer the weather occurs mainly in troposphere the most cloud formed in this layer troposphere is closed to the earth its depth varied according to the regions of the earth. Troposphere is more dipper in warmer region .

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3 years ago
A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.
alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Here,

k = Spring constant

m = Mass

Our values are given as,

m = 35g = 35*10^{-3}kg

f = 1 Hz

Rearranging to find the spring constant we have that,

k = (2\pi f \sqrt{m})^2

k = 4\pi^2 f^2 m

k = (4) (\pi)^2 (1) (35*10^{-3})

k = 1.38N/m

Therefore the spring constant is 1.38N/m

7 0
4 years ago
Please help me : (
Alex777 [14]
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5 0
4 years ago
A grandfather clock is controlled by a swinging brass pendulum that is 1.2 m long at a temperature of 27°C. (a) What is the leng
stealth61 [152]

Answer:

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

Explanation:

Given;

Initial length L1 = 1.2m

Initial temperature T1 = 27°C

Final temperature T2 = 0.0°C

Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C

The change i length ∆L;

∆L = L2 - L1

L2 = L1 + ∆L ...........1

∆L = xL1(∆T)

∆L = xL1(T2 - T1) ......2

Substituting the given values into equation 2;

∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)

∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)

∆L = -6.156 × 10^-4 m

From equation 1;

L2 = L1 + ∆L

Substituting the values;

L2 = 1.2 m + (- 6.156 × 10^-4 m)

L2 = 1.2 m - 6.156 × 10^-4 m

L2 = 1.1993844 m

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

3 0
3 years ago
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