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Nutka1998 [239]
3 years ago
8

What is the mass of an object that requires a force of 30 N to accelerate at a rate of 5 m/sec2 ?

Physics
1 answer:
Anastaziya [24]3 years ago
8 0
M=F/A
Which means 30 divided by 5 m/s is 6kg(mass)
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icang [17]

Answer:

Explanation:

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3 years ago
As a 5.0 x 102 N basketball player jumps from the floor up toward the basket, the magnitude of the force of her feet on the floo
s2008m [1.1K]
Here is your answer:

First find the notations:

5.0×10^2=500.
1.0×10^3=1,000

Then you multiply:

1,000×500=500,000

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=500,000 or 5.0×10^4
8 0
4 years ago
1. Describe the differences in the ways the sand castle is changed by an ocean wave and by Dylan stomping on it
Snezhnost [94]

A sand castle that was changed by the ocean waves would be slowly taken away from by the waves and would just become a smooth mound of sand. If Dylan stomped on it then it would still have all of the same sand as before but in a different shape. It would also become a mound of sand but it would have all the same sand as before.


(I don't know if I was supposed to talk about fancy physics stuff so I just gave a basic explanation. I don't know if my answer is really right or not... hope it helped though :))

6 0
3 years ago
How do you think the formation of new stars is related to supernovas and planetary nebulae?
kogti [31]
By God wanting them to be the same kinda
5 0
4 years ago
A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial
V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF=2.1\times 10^{-12} F

1pF=10^{-12} F

d=5.0 mm=5\times 10^{-3} m

1mm=10^{-3} m

a.Maximum flux

\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}

\epsilon_0=8.85\times 10^{-12}

\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm

b.Maximum displacement current,I=\frac{V}{R}=\frac{30}{130}=0.23 A

c.We have to find electric flux at t=0.5 ns

t=0.5ns=0.5\times 10^{-9} s

1ns=10^{-9}s

q=CV(1-e^{-\frac{t}{RC}})

q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})

q=52.9\times 10^{-12} C

\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm

d.Displacement current at t=0.5ns

I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}

I=0.037 A

5 0
3 years ago
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