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Lynna [10]
3 years ago
11

What net force is required to accelerate a car at a rate of 5 m/s ^ 2 if the car has a mass of 3, 000 kg ?

Physics
1 answer:
Olenka [21]3 years ago
3 0

Answer:

The answer is 15000kgms^-2

Explanation:

Actually here we are using the formula F=ma

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Which object would have the greatest acceleration?
Andre45 [30]

Answer:

D

Explanation:

A and C are balanced, B has a resultant force of 5N right, and D has a resultant force of 20N right.

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Given the equation E = P/N , solve for P
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Answer:

E=P/N

multiply both sides by N

P=EN

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sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o
crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

F_{net} = F_1 - F_2

F_{net} = 60 - 45

F_{net} = 15 N

so boat will have net force F = 15 N in forward direction due to both jumps

3 0
3 years ago
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aleksley [76]
The meters per second
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6 0
2 years ago
A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t
navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that v^2=u^2+2as

16^2=24^2+2\times a\times 50

a=-3.2m/sec^2

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So v^2=u^2+2as

0^2=24^2-2\times 3.2\times s

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration a=-3.2m/sec^2

So time t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec

7 0
3 years ago
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