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2 years ago

An object has moved through a distance. Can it have zero displacement? Justify your answer.PLEASE I WILL MARKK AS BRAINLIESTT

Physics

1 answer:

DochEvi [55]2 years ago

3 0

Answer:

Explanation:

yes object can move through a distance and can have zero displacement

as displacement is the shortest distance between two points eg: if a person has moved a distance of 20 m and comes back to his original place by moving back 20m in such a case distance moved by the person is 40m but his displacement is 0 as shortest distance covered by him is 0 as he is in his original place

Yes, it is possible for a body to have traveled a distance yet have a displacement of zero.

Distance traveled by a body is the length of the path covered by it.

On the other hand displacement of a body is the shortest distance between the initial and the final point of a body.

When a body is traveling such that it ends at the point from where it started its journey, it covers a distance but the displacement of the body is zero.

So a body that has moved through a distance can have zero displacement.

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3 years ago

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A ball rolls onto the path of your car as you drive down a quiet neighborhood street. To avoid hitting the child that runs to re

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Answer:

a) F = -1035.385 N

b) Backwards

c) s = 15.60 m

Explanation:

Given information

$u$ = Initial Speed of Car = 15.0 m/s

$v$ = Final Speed of Car = 9.00 m/s

$t_{b}$ = Breaking Time = 1.30 s

$m$ = Mass of Car = 1040 kg

Part (a)

To find the force exerted on the car we use the following formula

$F = ma$

Where

$F$ = Force = unknown

$m$ = Mass of Car = 1040 kg

$a$ = Acceleration of Car / Deceleration of Car = unknown

To find the force (F) we need to first find the deceleration rate (a)

To find the deceleration rate we use the following formula

$a = \frac{v - u}{t}$

Inputting the given values

$a = \frac{15 - 9}{1.30} \\ a = -4.615$

To find the force

$F = ma \\ F = (1040)(-4.615) \\ F = (1040)(-4.615) \\ F = -1035.385 N$

Part (b)

Since the value of F is negative this means the the force was opposite the direction of motion, hence the force was backwards.

Part (c)

To find the total distance the car moved while braking we use the following formula

$v^2 = u^2 + 2as$

Where

$s$ = distance traveled

Inputting the values given

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3 years ago

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

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