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3 years ago

HELP PLS 100 POINTS TAKING TEST RN

Physics

1 answer:

Basile [38]3 years ago

7 0

Answer:

sorry I hate when people press your question and answer with a fake answer for points, however you said it was 100 points and it's not even close lol but it is "your speed Is constant" that is not true

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The whole body metabolic rate of a bottlenose dolphin is 8000kcalsday. The dolphin weighs 190kg. What is the mass specific metab

Komok [63]

Answer: 42.1

Explanation:

Mass specific metabolic rate of a dolphin can be defined as the rate at which the dolphin consume energy per unit mass of body weight.

R = E/M

Where R = mass specific metabolic rate

E = Energy consumption = 8000kcalsday

M = mass = 190kg

R = 8000kcalsday/190kg

R = 42.1

4 0

4 years ago

In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas

Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

3 0

3 years ago

Read 2 more answers

A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.

Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

$v_f = \dfrac{872.10-851.10}{2}$

$v_f =10.5\ Hz$

v_o = 872.1 - 10.5

$V_0 = 861.6\ Hz$

The speed of jogger

$v = \dfrac{v_1 \times 343}{v_0}-343$

$v = \dfrac{872.1 \times 343}{861.6}-343$

v = 4.18 m/s

5 0

3 years ago

A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes

Ivan

Answer:

f = 19,877 cm and P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

1 / f = 1/97 + 1/25

1 / f = 0.05

f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

P = 1 / f

P = 1 / 19,877 10-2

P = 5D

5 0

3 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago