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gayaneshka [121]
3 years ago
10

Which statements below are true? Check all that apply. A. If motion starts and stops at the same location, then the displacement

is zero. B. Distance measurements must include magnitude, unit, and direction. I c. Distance is always greater than or equal to the magnitude of the displacement D. Displacement can be fully described with a magnitude and a unit.​
Physics
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Answer:

A. If motion starts and stop at the same location, then the displacement is zero

C. Distance is always greater than or equal to the magnitude of the displacement

Explanation:

The displacement of a body is a vector quantity that gives the shortest (straight line) distance moved by an object in motion, from its initial position to its final position. The displacement can also be described as the final position relative to the initial position of an object, such that the displacement vector, 's', is given as follows;

s = x_f - x _i

The displacement gives the direct length between two points, with regards to the direction of motion between the points. It can only be described by both the magnitude and the direction

Therefore, displacement can be positive (forward motion), negative (backward motion), or zero (where there is motion that stops at the starting point)

Therefore, a correct options is; <em>If motion starts and stop at the same location, then the displacement is zero</em>

<em />

Displacement, being a vector quantity will have both magnitude and direction

Distance, however, is a scalar quantity, it specifies only magnitude, but does not specify direction

Distance, describes the length of the path taken from one point to the other. Distance gives the total movement a body makes, without regards to the direction of the motion

Therefore, distance is always positive or zero (here only the object is stationary)

Therefore, a correct option is; <em>Distance is always greater than or equal to the magnitude of the displacement</em>

Therefore, the correct options are;

If motion starts and stop at the same location, then the displacement is zero

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A child moving at constant velocity carries a 2 N ice cream cone 1 m across a level surface. Is she doing work?
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Answer:

It would be A

Explanation:

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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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Read 2 more answers
At the instant a traffic light turns green, a car starts from rest with a given constant acceleration of 0.5 m/s squared. Just a
lubasha [3.4K]

Answer:

d= 1024 m

Explanation:

Kinematics of the car

The car moves with uniformly accelerated movement we apply the following formula:

d= v₀t+ (1/2)*a*t² Formula (1)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

a: acceleration in m/s²

Data

v₀₁= 0   :initial speed of the car

a₁ =0.5 m/s² : acceleration of the car

Kinematics equation of the car

We replace data in the formula (1) :

d= 0+ (1/2)*0.5*t²

d= (0.25)*t²  Equation (1)

Kinematics of the bus

The car moves with uniformly movement ( constant speed) we apply the following formulas:

d= v*t   Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

Data

v =  16 m/s

Kinematics equation of the bus

We replace data in the formula (2) :

d= 16*t   Equation (2)

<em>Problem development</em>

when the car passes the bus the elapsed time (t) and the distance (d) is equal for both.

Equation (1) = Equation (2)=d

(0.25)*t² =16*t  We divide both sides of the equation by t

(0.25)*t = 16

t= 16/(0.25)

t= 64 s

We replace t in the equation  (2)

d= 16*t

d= 16*(64)

d= 1024 m

<em />

5 0
3 years ago
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