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a_sh-v [17]
3 years ago
14

How much force is needed to accelerate at 66 kg skier 4m/s^2 ?​

Physics
1 answer:
Ray Of Light [21]3 years ago
7 0
E force needed to accelerate a 68 kilogram-skier at a rate of
1.2
m
s
2
is 81.6 Net forces.
Explanation:
Take the mass(68kg) and the acceleration of the skier(
1.2
m
s
2
) and multiply them together
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How many newton-meters are equal to 1600 joules
Salsk061 [2.6K]
1 Joule IS 1 newton-meter.
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3 years ago
An ideal step-down transformer has a turns ratio of 1/106. An ac voltage of amplitude 170 V is applied to the primary. If the pr
Leya [2.2K]

Answer:

I_2=0.8586A

Explanation:

From the question we are told that:

Turns Ratio i.e\frac{N_2}{N_1}=\frac{1}{106} (step down)

Voltage v=170v

Primary current amplitudeI_1= 8.10 mA =>8.10*10^{-3}A

Generally the equation for a Transformer is mathematically given by

\frac{E_1}{E_2}=\frac{I_2}{I_1}=\frac{N_1}{N_2}

Therefore

I_2=\frac{N_1}{N_2}*I_1

I_2=106*8.10*10^{-3}

I_2=0.8586A

7 0
2 years ago
The relation of mass m, angular velocity o and radius of the circular path r of an object with the centripetal force is-
k0ka [10]

Answer:

Correct option not indicated

Explanation:

There are few mistakes in the question. The angular velocity ought to have been denoted with "ω" and not "o" (as also suggested in the options).

The formula to calculate a centripetal force (<em>F)</em><em> </em>is

<em>F</em> = mv²/r

Where m is mass, v is velocity and r is radius

where

While the formula to calculate a centrifugal force (F) is

<em>F </em>= mω²r

where m is mass, ω is angular velocity and r is radius of the circular path.

<u>From the above, it can be denoted that the relationship been referred to in the question is that of a centrifugal force and not centripetal force, thus the correct option should be</u> C.

NOTE: Centripetal force is the force required to keep an object moving in a circular path/motion and acts inward towards the centre of rotation while centrifugal force is the force felt by an object in circular motion which acts outward away from the centre of rotation.

8 0
3 years ago
Forces and The Measurement of Work <br> match the following on the picture
sashaice [31]

Answer:

Explanation:

3 0
3 years ago
Read 2 more answers
A spring with a spring constant of 25.1 N/m is attached to different masses, and the system is set in motion. What is its period
Ratling [72]

1) 2.17 s

The period of a mass-spring oscillating system is given by

T=2 \pi \sqrt{\frac{m}{k}}

where k is the spring constant and m is the mass attached to the spring. In this problem, we have

k = 25.1 N/m

m = 3.0 kg

Substituting into the equation, we find

T=2 \pi \sqrt{\frac{3.0 kg}{25.1 N/m}}=2.17 s

2) 0.46 Hz

The frequency of the oscillating system is equal to the reciprocal of the period:

f=\frac{1}{T}

Therefore, by substituting T=2.17 s, we find:

f=\frac{1}{T}=\frac{1}{2.17 s}=0.46 Hz

3) 0.13 s

As before, the period of a mass-spring oscillating system is given by

T=2 \pi \sqrt{\frac{m}{k}}

where k is the spring constant and m is the mass attached to the spring. In this part of the problem, we have

k = 25.1 N/m

m = 11 g = 0.011 kg

Substituting into the equation, we find

T=2 \pi \sqrt{\frac{0.011 kg}{25.1 N/m}}=0.13 s

4) 7.69 Hz

The frequency of the oscillating system is equal to the reciprocal of the period:

f=\frac{1}{T}

Therefore, by substituting T=0.13 s, we find:

f=\frac{1}{T}=\frac{1}{0.13 s}=7.69 Hz

5) 1.59 s

Again, the formula for the period of a mass-spring oscillating system is given by

T=2 \pi \sqrt{\frac{m}{k}}

In this part of the problem, we have

k = 25.1 N/m

m = 1.6 kg

Substituting into the equation, we find

T=2 \pi \sqrt{\frac{1.6 kg}{25.1 N/m}}=1.59 s

6 0
3 years ago
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