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2 years ago

A constant force of 4.0 N acts on a variety of objects. The greatest acceleration will occur in the object that has

Physics

2 answers:

svlad2 [7]2 years ago

4 0

Answer:

Its the LEAST inertia

Explanation:

Trust me I pressed more inertia and got it wrong

aleksandr82 [10.1K]2 years ago

3 0

I think it will be the one with the more inertia

im not sure but hope this somewhat helped

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Debby and Ben took different routes to travel from Point A to Point E. Debby took the route along A, B, C, D, and E. Ben took th

svetlana [45]

Answer:

Ben's average speed was twice Debby's average speed.

Explanation:

Ben covered a total distance of 16 miles (10+4+2) and Debby covered 8 miles (3+2+2+1) which is half of what Ben covered. As they both reached the place in the same amount of time it tells us Ben was faster.

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3 years ago

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2 years ago

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

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2 years ago

A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up

prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)

We know that, Net force = m × a

$\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}$

$\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}$

$\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}$

$a_{\text {rad }}=2.45 \mathrm{m} / \mathrm{s}^{2}$

Now, Net force = m × a

Net force = 790 × 2.45

Net force = 1935.5 N

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3 years ago

A soccer player kicks a ball, applying a force of 1,000 newtons over a distance of 0. 2 meter. The ball travels 50 meters down t

Fed [463]

500 i think i’m wrong though

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2 years ago