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IRINA_888 [86]
2 years ago
14

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

the see saw in order to perfectly balance the child and rock?
Physics
1 answer:
ivann1987 [24]2 years ago
6 0

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock  

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

 (m)child(d)child)=(m)rock(d)rock)  

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

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An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

3 0
2 years ago
A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the b
Butoxors [25]

Answer:

v_{o}=18m/s

Explanation:

According to the exercise we know the angle which the bait was released and its maximum height

\beta =25\\y=2.9m

To find the initial y-component of velocity we need to do the following steps:

v_{y}^{2} =v_{oy}^{2}+2g(y-y_{o})

At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0

0=v_{oy}^{2}-2(9.8m/s^2)(2.9m)

v_{oy}=\sqrt{2(9.8m/s^{2} )(2.9m)} =7.53m/s

Since the bait is released at 25º

v_{oy}=v_{o}sin(25)

v_{o}=\frac{7.53m/s}{sin(25)}=18m/s

6 0
1 year ago
it possible that the net kinetic energy for two objects be zero while the net momentum is zero? Explain.
svp [43]
Of course. That's what you have when both objects are at rest. I'm guessing that you left a word out of the question, and it actually says that the net kinetic energy is NOT zero. In that case, the answer is still 'yes', but you have to think about it for a second.
4 0
2 years ago
An all female guitar septet is getting ready to go on stage. The lead guitarist, Kira,who is always in tune, plucks her low E st
matrenka [14]

Answer:

Buffy > Aiko > Chandra > Freja > Evita

Explanation:

Beats occur when two waves of nearby frequencies overlap. The number of beats per second is equal to the difference in the frequency.

Let

f= Frequency of Kira

f_1= Frequency of Aiko

The beat frequency of Kira and Aiko is therefore,

f_{beat}=|f_2-f_1|

Substituting  f  for  f_2  and 3Hz for f_{beat} , we get

3Hz=|f-f_1|\\f_1=f+3Hz

From the above equation, we see that beats increase

when

f_2 = Frequency of Chandra

The beat frequency of Kira and Chandra is therefore,

1Hz=|f-f_2|\\f_2=f-1Hz

From the above equation, we see that beats decrease

When

f_3= Frequency of Evita

The beat frequency of Kira and Evita is therefore,

5Hz=|f-f_3|\\f_3=f-5Hz

From the above equation, we see that beats decrease

When

f_4= Frequency of Freja

The beat frequency of Kira and Freja is therefore,

3Hz=|f-f_4|\\f_4=f-3Hz

From the equation above, we see that beats decrease

When

f_5= Frequency of Buffy

The beat frequency of Kira and Buffy becomes,

4Hz=|f-f_5|\\f_5=f+4Hz

From the equation above, we see that beats increase

Hence, the rank of members based on initial frequencies from largest to smallest is Buffy > Aiko > Chandra > Freja > Evita

6 0
1 year ago
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob
givi [52]

Answer:

Charge, Q=1.56\times 10^{-8}\ C

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

E=\dfrac{kQ}{r^2}

Q is the charge on an object

Q=\dfrac{Er^2}{k}

Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}

Q=1.56\times 10^{-8}\ C

So, the charge on the object is 1.56\times 10^{-8}\ C. Hence, this is the required solution.

7 0
2 years ago
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