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Cerrena [4.2K]
3 years ago
7

If you know the amount of air pressure exerted on a tabletop, how can you calculate the force exerted on the tabletop? Multiply

the air pressure by the area of the tabletop. Divide the air pressure by the area of the tabletop. Divide the area of the tabletop by the air pressure. Add the air pressure and the area of the tabletop.
Physics
1 answer:
babymother [125]3 years ago
7 0

Answer:

Multiply the air pressure by the area of the tabletop.

Explanation:

The relationship between pressure, force and area is given by:

p=\frac{F}{A}

where in this case, p is the air pressure, F is the force exerted and A the area of the tabletop. By re-arranging the equation, we can solve for F, the force exerted:

p=\frac{F}{A}\\p\cdot A=\frac{F}{A}\cdot A\\pA=F

So, the correct answer is:

The force exerted on the tabletop can be found by multiplying the air pressure by the area of the tabletop.

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Alchen [17]

Answer:

its the third one

Explanation:

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3 years ago
2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo
seropon [69]

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

7 0
3 years ago
a mass of 0.75 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 180 N/m, and t
Artemon [7]

Answer:

6.57 m/s

Explanation:

First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement

F=kx; F=180(.3) = 54 N

Next from Newton's second law find the acceleration of the mass.

Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

Now use the kinematic equation for velocity (or speed)

v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

v₀=0, since the mass is at rest before we release it

a=72 m/s² (from above)

x₀=0 as the start position already compressed

x₂=0.3m (this puts the spring back to it's natural length)

v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

v₂=\sqrt{43.2)\\ = 6.57 m/s

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A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
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Answer:

0.466 (3 sig. fig.)

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