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Shkiper50 [21]
2 years ago
7

Express force in terms of base units​

Physics
1 answer:
denis-greek [22]2 years ago
3 0

Answer:

F = [MLT⁻²]

Explanation:

Force = ma

m (mass) = [M]

a (acceleration) = [LT⁻²]

F(force) = m x a = [MLT⁻²]

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konstantin123 [22]

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4 0
3 years ago
A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical
Alex_Xolod [135]

Answer:

109.32 N/m

Explanation:

Given that

Mass of the hung object, m = 8 kg

Period of oscillation of object, T = 1.7 s

Force constant, k = ?

Recall that the period of oscillation of a Simple Harmonic Motion is given as

T = 2π √(m/k), where

T = period of oscillation

m = mass of object and

k = force constant if the spring

Since we are looking for the force constant, if we make "k" the subject of the formula, we have

k = 4π²m / T², now we go ahead to substitute our given values from the question

k = (4 * π² * 8) / 1.7²

k = 315.91 / 2.89

k = 109.32 N/m

Therefore, the force constant of the spring is 109.32 N/m

8 0
2 years ago
The strongest ernergy wave in the EM Spectrum is 'microwave'<br> true<br> false
Ratling [72]
Pretty sure it’s false....................
6 0
2 years ago
As you ride a bicycle on the sidewalk with a speed of
mixer [17]

Answer:

1.2 seconds

Explanation:

distance = ((final speed + initial speed) * time)/2

Here given:

  • distance: 3.8 meter
  • initial speed: 6.4 m/s
  • final speed: 0 m/s

Solving steps:

3.8 = ((0 + 6.4) * time))/2

3.8 = 3.2(time)

time = 3.8/3.2

time = 1.1875 seconds ≈ 1.2 seconds

5 0
1 year ago
A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed
GrogVix [38]

Answer:

  • a. \tau =  2.1161 s
  • b. V(18.8 \ s) = 0.0256 \ V

Explanation:

<h3>a.</h3>

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

V(t) = V_0 e^{(- \frac{t}{\tau}) }

where V_0 is the initial voltage, and \tau is the time constant.

For our problem, we know

V_0 = 185 \ V

and

V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

So

185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }

ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })

- \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })

\tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }

This gives us

\tau =  2.1161 s

and this is the time constant.

<h3>b.</h3>

At t = 18.8 s we got:

V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 0.0256 \ V

4 0
3 years ago
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