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Alexxx [7]
3 years ago
6

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

in 6.73s. what is the magnitiude of the girls average acceleration.
Physics
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

a = 0.53 m/s^2

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

\omega = 2\pi f

\omega = 2\pi ( \frac{20}{60})

\omega = 2.10 rad/s

so final tangential speed is given as

v = r\omega

v = 1.71 (2.10) = 3.58 m/s

now average acceleration of the girl is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{3.58 - 0}{6.73}

a = 0.53 m/s^2

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Sound waves cannot carry energy through. A water B air C a mirror D a vacuum
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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
3 years ago
Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between
Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

\Delta V=Ed

where

\Delta V is the potential difference

E is the magnitude of the electric field

d is the  distance between the plates

In this problem, we have:

\Delta V=750 V is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m

6 0
3 years ago
Does anyone know this
NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net  force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the  right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

  • Force to the right: F_a = 20 N, the applied force
  • Force to the left: F_f = 4 N, the force of friction
  • Force to the bottom: F_g = 400 N, the weight of the box (the weight is always downward vertically)
  • Force to the top: F_N = 400 N. This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

F_x = F_a - F_f = 20 - 4 = 16 N (to the right)

While the vertical force is

F_y = F_N - F_g = 400 - 400 = 0

So the net force is 16 N to the right.

2)

In this case, we have the following forces:

  • F_g = 4 N downward, the weight of the ball
  • F_a = 100 N to the left, the force that kicks the ball
  • F_f = 2 N to the right, the force of friction
  • F_N = 4 N upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

F_x =F_a - F_f = 100 -2 = 98 N (to the left)

While the vertical force is

F_y = F_g - F_N = 4 - 4 = 0 (downward)

And so, the net force is 98 N to the left.

3)

The force acting on the squirrel in this problem are:

  • F_g = 8 N downward, the weight of the squirrel
  • F_f = 7.5 N upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

F_y = F_g - F_f = 8 - 7.5 = 0.5 N

And since the weight is larger than the air resistance, the direction of the net force is downward.

4)

The forces acting on Monkey are:

  • F_1=95 N is the force applied to the right by Bunny
  • F_2 = 75 N is the force applied by Deer from the left (so, also on the right)
  • F_g = 50 N is the weight of Monkey, downward
  • F_N = 50 N is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

F_x = F_1 + F_2 = 95+75=170 N (to the right)

While the net force in the vertical direction is

F_y = F_N - F_g = 50 - 50 = 0

And therefore the net force is 170 N to the right

5)

The forces acting on Deer are:

  • F_a = 100 N + 100 N = 200 N to the right, the combined force applied by Bunny and Monkey
  • F_f = 25 N to the left, the force of friction
  • F_g = 150 N downward, the weight of the deer
  • F_N = 150 N upward, the normal reaction from the surface that balances the weight

So the net horizontal force is

F_x = F_a - F_f = 200 - 25 = 175 N to the right

While the net vertical force is

F_y = F_N - F_g = 150 - 150 = 0

So the net force is 175 N to the right.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

3 0
3 years ago
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