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3 years ago

Displacement can be zero but distance can't. explain this statement with example.

Physics

1 answer:

harkovskaia [24]3 years ago

5 0

Answer:

1}can displacement be zero even when distance is not zero? Give example.

2}how can you get speed of an object from its distance time graph ?

3}how can you get distance of an object from its speed time graph ?

4}can a body has zero velocity and still acceleration? give example .

5}is it possible in straight line motion of particle having zero speed and a non zero velocity? explain .

Explanation:

Hi friend,1).yes,displacement can be zero even when distance is not zeroexample :- when a body travels in circular path , after covering a circle the distance cannot be zero,but its displacement is zero..2).by finding the slope of distance time graph,we can find the speed of the body.3).by finding the integrating of speed time graph ,we can find the distance.4).yes,a body can have zero velocity and still acceleration.example- in vertically upward projected body,at its highest point,the velocity becomes zero but still acceleration due to gravity acts on it 5).no its not possible ..when the speed is zero..its velocity will be zero

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Acrostic poem for cell theory. Especially theory

Sonja [21]

Answer:

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Explanation:

The three tenets to the cell theory are as described below:

All living organisms are composed of one or more cells.

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3 years ago

Look at the distance-time graph for a remote-controlled car. What is the speed of the car?

Dafna1 [17]

Speed = distance/time

Speed = 50/5
Speed = 10m/s

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3 years ago

Read 2 more answers

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

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3 years ago

A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3

vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

$k = 5\:\text{N/m}$

The spring potential energy $PE_s$ is

$PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2$

$\:\:\:\:\:\:\:=22.5\:\text{J}$

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2 years ago

You are riding a bicycle. If you apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg, what wil

bekas [8.4K]

<h3>Hello There!!</h3>

<h3><u>Given</u>,</h3>

Force(F) = 150N

Mass(m) = 90kg

Acceleration(a) = ?

F= m×a

$150 = 90 \times \text{a} \\ \\ \text{a} = \frac{150}{90} \\ \fbox{cancelling by 3} \\ \\ \text{ a} = \cancel \frac{150}{90} \\ \\ \text{ a} = \frac{5}{3} = 1.67 \text{m/s} {}^{2}$

$\therefore \text{Option A= 1.67 m/s² is the correct answer}$

<h3>Hope this helps</h3>

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2 years ago

Displacement can be zero but distance can't. explain this statement with example.​

Displacement can be zero but distance can't. explain this statement with example.