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r-ruslan [8.4K]
3 years ago
6

If a force of 250N is applied to a piano being pushed up a ramp at an angle of 18degrees above the horizontal, what force is bei

ng applied to lift the piano?

Physics
1 answer:
klasskru [66]3 years ago
4 0

answer

77.25 N

explanation

(look at picture for free body diagram)

the force on the piano has both vertical and horizontal components, and since the questions asks about the force to lift the piano, we want to find the vertical component

we can use trigonometry where sinA = opposite/hypotenuse

A = 18 degrees

hypotenuse = 250 N

sin18 = opposite (vertical component) / 250

vertical component = sin18 * 250

= 77.25 N

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A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6
mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87  = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

n=\dfrac{m}{M}

M=Molar mass

n=\dfrac{m}{M}

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3 0
3 years ago
A testing instrument that's used to measure electrical signals in a circuit and display them as waveforms on a screen is called
disa [49]
A testing instrument that's used to measure electrical signals
in a circuit and display them as waveforms on a screen is called
an oscilloscope.

8 0
3 years ago
If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
3 years ago
A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows
anastassius [24]

Answer:

In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.

Explanation:

Hi there!

Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.

In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).

Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:

∑Fy = W + N = 0 ⇒ W = N

The weight is calculated as follows:

W = (mb + ms) · g

Where:

mb = mass of the bobsleder.

ms = mass of the sled.

g = acceleration due to gravity.

In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:

Fr = N · μ

Where:

N = normal force.

μ = kinetic friction coefficient.

Since N = W = (mb + ms) · g

Fr = (mb + ms) · g · μ

If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:

∑F = m · a

Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).

∑F = Fr =  (mb + ms) · g · μ =  (mb + ms) · a

(mb + ms) · g · μ =  (mb + ms) · a

Solving for "a":

g · μ = a

3 0
3 years ago
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Lena [83]

Answer:

The amplitude of vibration of string will increase due to which loudness of sound will increase

Explanation:

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This is known as resonance condition, so guitar will produce same frequency sound as that of frequency of string.

Now if the string is plucked with increasing level of force then it will increase the amplitude of vibrations of the string due to which the sound produced in the guitar will also be of same level.

So here we can say that amplitude and intensity of sound related as

I = kA^2

so on increasing amplitude the intensity will increase and hence it will produce loud sound

8 0
3 years ago
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