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3 years ago

I need help quick pls . Its either physical or chemical change !

Physics

1 answer:

Trava [24]3 years ago

8 0

Physical, chemical, chemical, physical, I’m pretty sure. Natural things, like you starting to breathe heavier are physical most of the time!!!

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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instan

gizmo_the_mogwai [7]

Answer:

The velocity just before hitting the ground is $v_f = 30 m/s$

Explanation:

From the question we are told that

The initial speed is $u = 10 m/s$

The final speed is $v = 30 \ m/s$

From the equations of motion we have that

$v^2 =u^2 + 2as$

Where s is the distance travelled which is the height of the cliff

So making it the subject of the the formula we have that

$s = \frac{v^2 - u^2 }{2a}$

Where a is the acceleration due to gravity with a value $a = 9.8m/s^2$

$s = \frac{30^2 - 10^2 }{2 * 9.8 }$

$s = 40.8 \ m$

Now we are told that was through horizontally with a speed of

$v_x =10 m/s$

Which implies that this would be its velocity horizontally through out the motion

Now it final velocity vertically can be mathematically evaluated as

$v_y = \sqrt{2as}$

Substituting values

$v_y = \sqrt{(2 * 9.8 * 40.8)}$

$v_y = 28.3 \ m/s$

The resultant final velocity is mathematically evaluated as

$v_f = \sqrt{v_x^2 + v_y^2}$

Substituting values

$v_f = \sqrt{10^2 + 28.3^2}$

$v_f = 30 m/s$

5 0

3 years ago

An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a con

antoniya [11.8K]

(a) 4.03 s

The initial angular velocity of the wheel is

$\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s$

The angular acceleration of the wheel is

$\alpha = -3.40 rad/s^2$

negative since it is a deceleration.

The angular acceleration can be also written as

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 0$ is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

$t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s$

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

$\theta= \omega_i t + \frac{1}{2}\alpha t^2$

where we have

$\omega_i=13.7 rad/s$ is the initial angular velocity

$\alpha = -3.40 rad/s^2$ is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

$\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad$

6 0

3 years ago

A toy spacecraft is launched directly upward. When the toy reaches its highest point, a spring is released and the toy splits in

Murljashka [212]

Answer:

Explanation:

Momentum conservation will cause 0.08kg to move to the west (opposite of 0.02 kg).

and because both are at the same height above the ground, they will take the same time to reach the ground.

The speed of 0.08kg will be less than 0.02 kg, let v be the speed of 0..02kg, then speed of 0.08kg V is

0.02v - (0.08)V = 0

V = 0.02 v/ 0.08 = v/4

The speed of 0.08 kg = v/4

The speed of 0.08 kg is less than 0.02kg.

So 0.02kg strikes the ground farther from the launch point than does the 0.08 kg

8 0

3 years ago

A force of 20 N produces an acceleration of 10 m/s² in mass m1 and an acceleration of 5 m/s² in

Scrat [10]

Explanation:

F = 20N m= m1 a=10m/s²

m=m2 a=5m/s²

F = ma

for the first one: 

f=m1 × a

20 = m1 ×10

20=10m1

m1=20/10

m1=2

for the second one :

f=m2×a

20=m2×5

m2= 20/5

m2= 4

since F=ma

F=(m1+m2) ×a

F =(4+2)×a

F =6×a

F=20(from the question above )

20=6×a

a=20/6

a=3.33

8 0

4 years ago

Read 2 more answers

A race card drives one lap around a race track that is 500 meters in length. How is this displacement different from the distanc

gulaghasi [49]

Answer:

Distance is 500 m, displacement is 0

Explanation:

Distance and displacement are defined in two different ways:

- Distance is the total length of the path covered by an object in motion - so it depends on the path taken. In this problem, the distance travelled by the car corresponds to the length of one lap, which is the length of the track, so 500 m

- Displacement is the distance in a straight line between the final point and the initial point of the motion. This means that displacement does not depend on the path taken, but only on the starting and ending point of the motion. In this problem, the car completes one lap, so the final position of the car is equal to its starting position - therefore the displacement is zero, since the distance between these two points is zero.

5 0

3 years ago