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3 years ago

This question that it pictured

Physics

1 answer:

MrRissso [65]3 years ago

8 0

Work = (force) x (distance)

= (20 N) x (5 m)

= 100 joules

You really ought to be able to answer this kind of question by now.

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Andre the Giant and Thumbelina are standing in the middle of an ice-skating rink facing each other. Andre the Giant pushes Thumb

IceJOKER [234]

I don't know for sure but i think that it is because the force him pushing her pushed him back as well

Hope i helped

3 0

3 years ago

Read 2 more answers

The velocity of the water in the pipe at right is given by V1 = 0.5t m/s and V2 = 1.0t m/s, where t is in seconds. Determine the

Nonamiya [84]

Answer:

A) At point 1, local acceleration = 0.5 m/s²

At point 2, local acceleration = 1.0 m/s²

B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

Explanation:

Local acceleration at those points is the instantaneous acceleration at those points and it is given as

a = dv/dt

At point 1, v₁ = 0.5 t

a₁ =dv₁/dt = 0.5 m/s²

At point 2, v₂ = 1.0 t

a₂ = dv₂/dt = 1.0 m/s²

b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time

Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t

Time = t

Average acceleration = 0.5t/t = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

8 0

3 years ago

Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in

monitta

Person-1

Initial velocity=u=0m/s
Final velocity=v=10m/s
Time=10s=t

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}$

$\\ \sf\longmapsto Acceleration=\dfrac{10}{10}$

$\\ \sf\longmapsto Acceleration=1m/s^2$

Person-2

initial velocity=0m/s=u
Final velocity=v=0.25m/s
Time=t=2s

$\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}$

$\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}$

$\\ \sf\longmapsto Acceleration=0.125m/s^2$

Person-1 is accelerating faster.

4 0

3 years ago

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

$E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2$

$E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2$

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0

3 years ago

a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe

Mariana [72]

Answer:

1.34352 kg

Explanation:

$m_w$ = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

$\Delta T$ = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

$m_v$ = Mass of water in the vessel

Here the potential energy will balance the internal energy

$m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg$

Mass of the water in the vessel is 1.34352 kg

6 0

3 years ago