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3 years ago

How might you describe the mathematical procedure of finding the displacement when an object travels in two opposite directions?

1 answer:

4 0

Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.

Displacement = 2 km - 2km = 0

Generally, the equation is:
Displacement = Distance of motion to the right - Distance of motion to the left

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The amount of matter in an object is called its

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The amount of matter in an object is called its, mass. How much the mass weights would be referred to as weight

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3 years ago

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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 

Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
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3 years ago

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Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t

elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0

h = 1/2 gt²

$t = \sqrt{\frac{2h}{g} }$

Horizontal distance covered = t x Vx

= $\sqrt{\frac{2h}{g} } \times V_x$

From this expression also

Horizontal distance covered is proportional to $\sqrt{h}$ .

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3 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$