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crimeas [40]
2 years ago
13

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on th

e left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground? 
a. 4m
b. 5m
c. 2m
d. 3m​
Physics
1 answer:
Lapatulllka [165]2 years ago
4 0

Answer:

Option A. 4 m

Explanation:

Please see attached photo for diagram.

In the attached photo, X is the distance from the centre to which the student on the right must sit in order to balance the seesaw.

Clockwise moment = X × 45

Anticlock wise moment = 3 × 60

Clockwise moment = Anticlock wise moment

X × 45 = 3 × 60

X × 45 = 180

Divide both side by 45

X = 180 / 45

X = 4 m

Thus, the student must sit at 4 m from the centre.

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Infer why the doppler effect can or can not occur in all waves.
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Answer:

it can occur in all waves because all wave have a frequency

8 0
2 years ago
How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magne
ser-zykov [4K]

Answer:

The rate at which power is generated in the coil is 10.24 Watts

Explanation:

Given;

number of turns of the coil, N = 160

area of the coil, A = 0.2 m²

magnitude of the magnetic field, B = 0.4 T

time for field change = 2 s

resistance of the coil, R =  16 Ω

The induced emf in the coil is calculated as;

emf = dΦ/dt

where;

Φ is magnetic flux = BA

emf = N (BA/dt)

emf = 160 (0.4T x 0.2 m²)/dt

emf = 12.8 V/s

The rate power is generated in the coil is calculated as;

P = V²/ R

P = (12.8²) / 16

P = 10.24 Watts

Therefore, the rate at which power is generated in the coil is 10.24 Watts

8 0
3 years ago
 what is the difference between repelling and attracting
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Answer:

Attracting means pulling toward you and repelling means pushing away

Explanation:

3 0
2 years ago
Read 2 more answers
(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg
valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be m_ 1 = 80 kg

The weight of the buddy be m_2 = 120 kg

The weight of  Bubba be m_3 = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2

The length of the boat be x_2 = 4 m

∴

We can find the center of mass of the system by using the formula:

X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}

4 0
3 years ago
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