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scoray [572]
2 years ago
6

According to newton’s first law, when the net force acting on an object is zero, the object must _____.

Physics
1 answer:
zysi [14]2 years ago
5 0
Remain at rest.

Hope this helps you.
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For most people, getting 8 to 9 hours of sleep increases concentration, improves physical health, and improves one's mood.
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On the surface of Earth, the force of gravity acting on one kilogram is: *
Murrr4er [49]

Your answer would be:

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4 0
3 years ago
To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
8 0
3 years ago
A particle having a speed of 0.87c has a momentum of 10-16 kg·m/s. what is its mass?
cupoosta [38]
The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
p=mv
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m= \frac{p}{v} =3.8\cdot10^{-25}~Kg
8 0
3 years ago
When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity \rightarrow 20m/s

\lambda = Wavelength \rightarrow 35*10^{-2}m

Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

The frequency is directly proportional to the angular velocity therefore

\omega = 2\pi f

\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

Where

A = Amplitude

Then replacing,

V_{max} = (1*10^{-2})(359.03)

V_{max} = 3.59m/s

Therefore the maximum speed of a point on the string is 3.59m/s

8 0
3 years ago
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