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Sholpan [36]
2 years ago
11

What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0

Physics
1 answer:
ANEK [815]2 years ago
8 0

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

Initial velocity = 0

Time = 3 seconds.  

Using the Formula,  

Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,  

Force = mass × acceleration

= 20 × 1.6

=

Explanation: I REALLY  HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]

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Answer:

D

Explanation:

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7 0
3 years ago
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Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo
Bingel [31]

Answer:

t=14.678\times 10^{-3}s

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

Q=Q_o(1-e^{-\frac{t}{RC}})

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})

or

0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}

or

e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1

taking log both sides, we get

ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})

or

-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)

or

t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}

or

t=14.678\times 10^{-3}s

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What is the current in a 10V Circuit if the resistance is 2 ohms
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