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scZoUnD [109]
2 years ago
14

A 0.15 kg project hits the ground with a speed of 11 m/s. The project comes to rest in 0.015 seconds. What is the net force?

Physics
1 answer:
Blababa [14]2 years ago
4 0

Answer:

Explanation:

ACCORDING TO NEWTONS SECOND LAW;

F = mass * acceleration

F = m(v-u/t)

m is the mass = 0.15kg

v is the final velocity = 11m/s

u is the initial velocity = 0m/s

t is the time = 0.015

Substitute;

F = 0.15(11-0)/0.015

F = 0.15(11)/0.015

F = 1.65/0.015

F = 110N

Hence the net force is 110N

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A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge
Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b  = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c  = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π×10^{-7}×0.024 / 2π(1.20×10^{-2}) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 ×10^{-7}× 1.09861 × 120

current = 52.73 ×10^{-6}  A

so here current in loops is 52.73 μA

8 0
3 years ago
A 4.0 kg shot put is thrown with 30 N of force. What is its acceleration?
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7.5 m/s

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a = 7.5 m/s2

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