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PilotLPTM [1.2K]
4 years ago
8

As water changes state,the mass of the sample does not change.the size of the water molecules_______

Physics
1 answer:
Mamont248 [21]4 years ago
7 0

Answer:

does not change

Explanation:

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A rock thrown with speed 7.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 18.0 m bef
Iteru [2.4K]

Answer:

height from where rock was thrown is 27.916 m

Explanation:

speed = 7.50 m/s

θ = 30°

g= 9.8 m/s²

horizontal distance = 18 m

time require for vertical displacement

time = \frac{distance}{velocity} \\t = \frac{18}{7.5\ cos30^0}

t = 2.8 sec

now for calculation of height

s = ut + 0.5 a t²

-h = v sinθ× t + 0.5 ×(-9.8)× (2.8²)

-h = 7.5 sin30°× 2.8 - 0.5 ×(9.8)× (2.8²)

-h = -27.916 m

h= 27.916 m

height from where rock was thrown is 27.916 m

5 0
3 years ago
The bob (weight) at the end of a pendulum has a mass of 0.3 kilograms. The bob is pulled to position B and allowed to swing. It
Ivahew [28]

Answer:

0.147 J

Explanation:

The total energy that has been transformed into thermal energy is equal to the loss of gravitational potential energy between the initial situation (bob at h=0.5 m above the ground) and the final situation (bob back but at h=0.45 m above the ground).

Therefore, we have

E_{thermal}=\Delta U=mgh_1 - mgh_2 = mg(h_1 -h_2)

where

m = 0.3 kg is the mass of the bob

g = 9.8 m/s^2

h1 = 0.5 m is the initial height

h2 = 0.45 m is the final height

Substituting, we find the thermal energy

E_{thermal}=(0.3 kg)(9.8 m/s^2)(0.5 m-0.45 m)=0.147 J

Therefore, the energy transformed into thermal energy is 0.147 J.

3 0
3 years ago
hich answer represents a speed, not a velocity? A. 35 m/s east B. –35 m/s down C. 35 m/s D. 35 m/s south
finlep [7]
The answer is C. 35m/s because there is no direction

3 0
3 years ago
Read 2 more answers
Difference between force and acceleration
galina1969 [7]
Force your doing it purposely and acceleration it’s just happening
7 0
3 years ago
Read 2 more answers
A student that jumps a vertical height of 50 cm during the hang time activity.
muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

<u>Initial speed calculations:</u>

v_{f}^{2}=v_{o}^{2}+2*a*d

At max height, the speed tends to zero.

So, calculating, we have:

<u>v_{f}^{2}=v_{o}^{2}+2*a*d\\\\0=v_{o}^{2}+2*(-9.81\frac{m}{s^{2}})*0.5m\\\\v_{o}^{2}=9.81\frac{m^{2} }{s^{2}}\\\\v_{o}=\sqrt{9.81\frac{m^{2} }{s^{2}}}=3.13\frac{m}{s}</u>

<u>Hang time calculations:</u>

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s

Then, for the total hang time, we have:

TotalHangTime=2*0.318seconds=0.64seconds

Have a nice day!

3 0
4 years ago
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