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3 years ago

A seesaw pivots as shown in below. What is the net torque about the pivot point?

Physics

1 answer:

rodikova [14]3 years ago

7 0

Answer:

2.3 Nm clockwise

Explanation:

Take counterclockwise to be positive and clockwise to be negative.

∑τ = (3 N) (2.5 m) − (7 N) (1.4 m)

∑τ = 7.5 Nm − 9.8 Nm

∑τ = -2.3 Nm

The net torque is 2.3 Nm clockwise.

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A transformer is to be used to provide power for a computer disk drive that needs 6.0 V (rms) instead of the 120 V (rms) from th

Ipatiy [6.2K]

Answer:

$N_{2}$ =20 turns

Explanation:

The given case is a step down transformer as we need to reduce 120 V to 6 V.

number of turns on primary coil N_{P}= 400

current delivered by secondary coil I_{S}= 500 mA

output voltage = 6 V (rms)

we know that

$I_{p}=\frac{V_{out}}{V_{in}\times I_{s}}$

putting values we get

$I_{p}=\frac{6}{120\times 0.5}$

$I_{p}= 0.1 A$

to calculate number of turns in secondary

$\frac{N_{2}}{400} =\frac{6}{120}$

therefore, $N_{2}$ =20 turns

5 0

3 years ago

What energy is directly dependent upon velocity and mass?

erastova [34]

Answer:

Kinetic energy

Explanation:

3 0

2 years ago

You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient

iren [92.7K]

Answer:

angle minimum θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

N - W = 0

N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

the friction force formula is

fr = μ N

fr = μ W

we substitute

μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0

μ sin θ - cos θ + ½ cos θ= 0

μ sin θ - ½ cos θ = 0

sin θ / cos θ = 1/2 μ

tan θ = 1/2 μ

θ = tan⁻¹ (1 / 2μ)

θ = tan⁻¹ (1 (2 0.57))

θ = 41.3º

7 0

2 years ago

A mass m is attached to a string connected to a force sensor on a rotating platform. The platform’s angular velocity, ω, can be

makkiz [27]

Answer:

The mass m is 0.332 kg or 332 gm

Explanation:

Given

The platform is rotating with angular speed , $\omega =8.5\, \frac{rad}{sec}$

Mass m is moving on platform in a circle with radius , $r=0.20\, m$

Force sensor reading to which spring is attached , $F=4.8\, N$

Now for the mass m to move in circle the required centripetal force is given by $F=m\omega ^{2}r$

=> $4.8=m\times 8.5 ^{2}\times 0.20$

$=>m=0.332\, kg$

Thus the mass m is 0.332 kg or 332 gm

7 0

2 years ago

Please help i’ll mark you

suter [353]

Answer:

What should I do. reply quickly for a quick answer

8 0

1 year ago

Read 2 more answers