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kozerog [31]
3 years ago
5

This is due by 11:59 PM tonight.

Physics
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

1. increases

2. increases

3. increases

Explanation:

Part 1:

First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:

F1 - fs = 0.

And this friction force fs is:

fs = Nμs,

where μs is the static coefficient of friction, and N is the normal force.

Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:

N = mg + F2.

So, F2 is increasing, that means fs is increasing too.

Part 2:

As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.

Part 3:

In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.

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two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change
Verdich [7]
The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
5 0
3 years ago
A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has
lara [203]

Answer:

The phase constant is 7.25 degree  

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = \sqrt{(\frac{v_x}{\omega })^2+y^2 }  = \sqrt{(\frac{35}{2\times \pi  \times y})^2+6.2^2 }  =  6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ  = 0.992

Ф = 7.25 degree  

so the phase constant is 7.25 degree  

5 0
3 years ago
At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values
denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2

but we know that at resonance X_L=X_C  

putting  X_L=X_C in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}  so from given expression it is clear that quality factor depends on R L and C

3 0
3 years ago
A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16
erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle \phi = 20°

No. of teeth T = 17

Face width (b) = 20 mm

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Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

V =\pi D( \frac{N}{60} )

V = 3.14 (0.034) \frac{(1650)}{60}

V = 2.93 \frac{m}{s}

Form factor for the pinion gear is

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Now the bending stress is given by the formula

\sigma = \frac{K_{v} F}{m b y}

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8 0
3 years ago
Which of these surface features are formed because of the heating of groundwater?
Tju [1.3M]
The correct option is C.
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8 0
3 years ago
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