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3 years ago

The distance between an object and a reference point is the object's what?

Physics

2 answers:

timofeeve [1]3 years ago

7 0

Answer:

Displacement

Explanation:

The distance between an object and a certain reference point is called displacement. This is not to be confused with distance, which represents the total distance the object has travelled to get a certain point, regardless of how long it is. Displacement always represents the shortest distance (a straight line) between the initial and final points.

Komok [63]3 years ago

6 0

Answer:

Displacement is the distance and direction of an object's change in position from a reference point. Suppose a runner jogs to the 50-m mark and then turns around and runs back to the 20-m mark.

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A cadillac escalade has a mass of 2 569.6 kg, if it accelerate at 4.65m/s2 what is the net force of the car

anygoal [31]

Force is equal to mass multiplied by acceleration, therefore
F=ma
m=2569.6 kg
a=4.65m/s^2
therefore F=2569.6*4.65=11948.6 (correct to 1 d.p.)

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3 years ago

All of the following were barriers to minority participation in early psychology except

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2 years ago

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While in empty space, an astronaut throws a ball at a velocity of 13 m/s. What will the velocity of the ball be after it has tra

aliya0001 [1]

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3 years ago

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A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For maximum emf, $\sin\omega t=1$

So,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

So, the maximum emf generated in the loop is 5.32 V.

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3 years ago

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little

horrorfan [7]

Answer:

t_up / t_down = 6.83

Explanation:

Find:

Calculate the ratio of the time he is above y_max/2 to the time it takes him to go from the floor to that height.

Solution:

- Compute the velocity v_o at y_max:

v_i^2 = v_f^2 - 2*g*y_max

0 = v_o^2 - 2*g*y_max

v_o = sqrt (2*g*y_max)

- The total time spend by athlete above height y_max / 2 is:

y - y_o = v_o*t_up - 0.5*g*t^2_up

v_o = 0.5*g*t_up

- Equate two equations:

sqrt (2*g*y_max) = 0.5*g*t_up

t_up = 2*sqrt(2*g*y_max) / g

- The total time taken by athlete to reach height y_max / 2 from ground is:

y - y_o = v_o*t_down - 0.5*g*t^2_down

g*t_^2down - 2*v_o*t_down + y_max = 0

- Solve the quadratic and evaluate t_down:

t_down = (v_o +/- sqrt (v^2_o - g*y_max)) / g

Substitute for v_o = sqrt (2*g*y_max)

t_down = (sqrt(2g*y_max) +/- sqrt(g*y_max)) / g

- We will use the minus quantity, because we need the first part of the journey from ground from the two times he passes the height of y_max/2.

Hence,

t_down = (sqrt(2g*y_max) - sqrt(g*y_max)) / g

t_down = (sqrt(g*y_max) / g) * (sqrt(2) - 1)

- Compute the ratio t_up to t_down:

t_up / t_down = 2*sqrt(2*g*y_max) / g * g / (sqrt(g*y_max)*(sqrt(2) - 1)

= 2*sqrt(2) / (sqrt(2) - 1)

= 6.83

8 0

2 years ago