Ondas de agua en un plato poco profundo tienen 6

Anyone going to be my friend

Artemon [7]

Explanation:

I'd love to but we cant talk right now cause its 12:22 am here and I'm gonna sleep now lol.

but let's follow each other.

who knows we might be able to help each other.

whaddya say?

have a good day ♡

5 0

3 years ago

The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?

Talja [164]

Answer:

75 rotations

Explanation:

f0 = 0, f = 3000 rpm = 50 rps, t = 3 s

(a) use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 50 = 0 + α x 3

α = 104.67 rad/s^2

(b) Let θ be the angular displacement

use second equation of motion for rotational motion

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 104.67 x 3 x 3

θ = 471.015 rad

The angle turn in one rotation is 2 π radian.

Number of rotation = 471.015 / (2 x 3.14) = 75 rotations

7 0

3 years ago

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$

$V_{x} = -1.40(7.30) + 22$

$V_{x} = -10.22 + 22$

$V_{x} = 11. 78 m/s$

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

Area of triangle + Area of rectangle

$[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

= 37.303 + 85.994

= 123. 297 m

≈ 123. 30 m

4 0

3 years ago

Read 2 more answers

A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0

3 years ago

Suppose that a person riding on the top of a freight car shines a searchlight beam in the direction in which the train is travel

Snezhnost [94]

Answer:

same

Explanation:

Acc. to Einstien's postulate of special theory of

Relativity , Velocity of the light beam is same in all frames of references

(a) If the freight car is at rest

The frame we can assumed as Non - inertial frame of reference s

In the inertial frame of reference , velocity of the light beam has its own value as : 3 x 10^8 m/s

(b) If the freight car is moving , the frame we can assumed as Non -inertial frame of reference

In thus case also , The velocity of the light beam will also have the same value as ; 3 x 108 m/s

6 0

3 years ago