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3 years ago

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An object with a mass of 5.0 kg accelerates 2.8 m/s2 towards right when an unknown force is applied to it. What is the amount of

the force? What is the direction of force?

1 answer:

RoseWind [281]3 years ago

5 0

Answer:

4n

Explanation:

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(100, 108)
Due to
1.2x90=108

100, 108

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____ are solid while lines stretching across one or more lanes in the same direction, indicating the proper place to come to a s

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Stop lines are solid white lines painted across the traffic lanes at intersections and pedestrian crosswalks, indicating the exact place to stop.

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3 years ago

The amplitude of a sound is the A. frequency of the sound. B. magnitude of displacement of a sound pressure wave. C. psychologic

AlladinOne [14]

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Option B. magnitude of displacement of a sound pressure wave

Explanation:

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3 years ago

Read 2 more answers

A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall

kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

$F = q(\vec v \times \vec B)$

so here we have

$F = qvB sin90$

since force is perpendicular to the velocity so here it must be centripetal force

here we have

$\frac{mv^2}{R} = qvB$

so we have

$R = \frac{mv}{qB}$

$R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}$

$R = 4.06 cm$

so here it will move in circle with radius 4.06 cm

8 0

3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

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3 years ago