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Ksju [112]
3 years ago
11

Recall: earth applies _____________ on Earth or equivalently

Physics
1 answer:
Olin [163]3 years ago
4 0

Answer:

gravitational attraction

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How can resistance be useful and where can they be useful?
sergiy2304 [10]
-- Resistance can be useful among the population of a repressive government.
Although it can be dangerous for those who resist, it can also exert pressure
against the regime to alter its repressive practices.

-- Resistance can also be useful in electronic circuits. "Lumped" components with
known numerical values of resistance are used to divide voltage, limit current, and
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2 years ago
Can I improve the design of my simple machine? How?
blsea [12.9K]

Answer:

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Explanation:

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5 0
3 years ago
A child on a skateboard experiences a 75 N force with an acceleration of 1.5 m/s2.
cupoosta [38]

Answer:

The mass of the child + skateboard is 50 kg

Explanation:

In this problem, we can apply Newton's second law:

F = ma

where

F is the net force on a system

m is the mass of the system

a is the acceleration of the system

In this problem, our system is the child + the skateboard. The net force on them is

F = 75 N

and their acceleration is

a=1.5 m/s^2

So we can re-arrange the equation above to find their combined mass:

m=\frac{F}{a}=\frac{75}{1.5}=50 kg

3 0
2 years ago
A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o
Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

             V_{average} = 4 / 2)

             V_{average} = 2 V

6 0
2 years ago
A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir
TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

I=I_0 cos^2\theta

Where,

I_ {0} indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

\theta indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)

Replacing the values,

I=100*cos^2(20)*cos^2(40-20)

I=100*cos^4(20)

I=77.91W/m^2

Therefore the intesity of the light after it has passes through both polarizers is 77.91W/m^2

7 0
2 years ago
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