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nalin [4]
3 years ago
12

Three of your friends are all sitting g on one end of a seesaw. The combined weight is 275 N. The length from the fulcrum to you

r friends is 2.5 m. The rest of the seesaw (from the fulcrum to you) is 4.5 m. What is the MA? What effort force is needed to lift your friends?
Physics
1 answer:
yanalaym [24]3 years ago
4 0

Answer:

1.8

153 N

Explanation:

MA = effort arm / load arm

MA = 4.5 / 2.5

MA = 1.8

MA = load force / effort force

1.8 = 275 N / F

F ≈ 153 N

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Explanation:

Q1) What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?

Ans1) speed, v=st=2πrT=2×227×7×10-260×60=119×10-4=1.22×10-4m/s

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Which component of a galaxy is often found between the stars and looks like a cloud or smoke?
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Mr. Fuller assigned his science class a lab comparing the masses of objects to the nearest gram using a balance. The mass of a p
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3 years ago
. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial lengt
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Answer:

a)    L = 3.29 10⁻⁴ H,  b)U = 5.33 10⁻²  J

Explanation:

a) The inductance is a solenoid this given carrier

           L = \frac{N \ \phi_B }{I}

The magnetic field inside the solenoid is

          B = μ₀ \frac{N}{l}  i

hence the magnetic flux

          Ф_B = B. A = μ₀ \frac{N \ A}{l \ i}

we substitute in the expression of inductance

          L = N² μ₀ A /l

let's find the area of ​​each turn

          A = π r²

         A = π 0.02²

         A = 1.2566 10⁻³ m²

let's calculate

          L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3

          L = 3.29 10⁻⁴ H

b) The stored energy is

           U = ½ L i²

let's calculate

            U = ½ 3.29 10⁻⁴ 18²

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5 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

7 0
3 years ago
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