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OlgaM077 [116]
3 years ago
13

4. You make the following measurements of an object: 42 kg and 22 m3. What would the object’s density be?

Physics
1 answer:
White raven [17]3 years ago
3 0
The formula for getting the density of an object is
      density = mass/volume =  ρ = m/v

      mass =42 kg
      volume = 22m3
we will work on the equation directly because they are already in there SI units
                          density = 42 kg/22m3
                          density = 1.9090909090909
which if rounded can become  1.9 kg/m3
      ans = the object's density is 1.9090909090909 kg/m3 or 1.9 kg/m3
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<u> Answer </u>

The impulse on the second trial is smaller is smaller than in the first trial.

<u>Explanation </u>

Impose of a body is that change in momentum during a time interval. If the change of momentum takes longer then, the impulse of a force is less. I a moving object hits a hard surface the rate of change of momentum is very high. e.i in the first trial, the egg breaks because it hits the hard surface(ground).

In the second trial, the foam cushion absorbs the shock and prolongs the time of impact with the egg hence decreasing the impulse.


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3 years ago
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As a way of determining the inductance of a coil used in a research project, a student first connects the coil to a 5.62 V batte
Reptile [31]

Answer:

Its inductance L = 166 mH

Explanation:

Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR

R = V/I = 5.62/0.698 = 8.052 Ω

Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from

V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'

Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω

WE now find the reactance X of the coil from

Z² = X² + R²

X = √(Z² - R²)

= √(97.5² - 8.05²)

= √(9506.25 - 64.8025)

= √9441.4475

= 97.17 Ω

Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.

L = X/2πf

= 97.17/2π(93.1 Hz)

= 97.17 Ω/584.965 rad/s

= 0.166 H

= 166 mH

Its inductance L = 166 mH

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3 years ago
What is the internal resistor of the cell in closed circuit?
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3 years ago
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,
Ne4ueva [31]
A)  f = 1.8 rev/s = 2 Hz 
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5 0
3 years ago
Can someone solve this problem and explain to me how you got it​
evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

4 0
3 years ago
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