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eimsori [14]
3 years ago
5

Which of the following is a field force

Physics
1 answer:
IRINA_888 [86]3 years ago
8 0
So basically my morning has been good so far I woke up with a. Headache tho maybe a I have a tumor idk
You might be interested in
A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti
aleksley [76]

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ =  30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

7 0
3 years ago
An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull
puteri [66]

Answer:

\mu=0.0049Kg/m

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

f_n=\frac{nv}{2L}

Where L is the length of the string and v the velocity of propagation. Use this expression to find the value of v.

f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s

The velocity of propagation is given by the expression:

v=\sqrt{\frac{T}{\mu }

Where \mu is the desirable variable of the problem, the linear mass density, and T is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

T=W=mg=(5)(9.81)=49.05N

With the value of the tension and the velocity you can find the mass density:

v=\sqrt{\frac{T}{\mu}

v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m

6 0
3 years ago
The rate at which a candle burns in millimeters per minute is:
anzhelika [568]
since both components, length and time, are measurable 
<span>since Rate = length ÷ time </span>
<span>∴ rate is also measurable and ∴ quantitative.

</span>
3 0
3 years ago
A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg
ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

p=m/V\\\\ p=m/(L*A)\\p*A=m/L

Now the equation for speed of traverse wave is calculated through:

\sqrt \frac{F*L}{m}\\

=\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}

Substituting the values

\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3}  \\

=22.49 m/s

4 0
3 years ago
How much force is required to raise a 0.2 kg mass?
Firlakuza [10]

Answer:

1.96 N

Explanation:

6 0
2 years ago
Read 2 more answers
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