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SSSSS [86.1K]
3 years ago
8

O ln a hydraulic press , a force Of 400 N is exerted on

Physics
1 answer:
egoroff_w [7]3 years ago
7 0

Explanation:

Pressure = force / area

P = 400 N / 0.001 m²

P = 400,000 Pa

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You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin
faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance L=0.450H=0.45\times 10^{-3}H

Current in the inductor i=1.90mA=1.90\times 10^{-3}A

Voltage v = 13 volt

Inductive reactance of the circuit X_l=\frac{v}{i}

X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm

We know that

X_l=\omega L=2\pi fL

2\times 3.14\times  f\times 0.45\times 10^{-3}=6842.10

f = 2421.127 kHz

6 0
4 years ago
A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an
Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

<u>ΔP = 0.056 psi</u>

8 0
4 years ago
What are some examples of aerobic exercise?
Firdavs [7]

Answer:

swimming, cycling, and jogging

Explanation:

4 0
3 years ago
A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter
icang [17]

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

7 0
3 years ago
Un objeto de 1400 g de masa se mueve bajo la acción de una fuerza constante con una aceleración de 0,5 m/s2 , sobre una superfic
igomit [66]

Answer:

if you spoke this in english i can help you out

Explanation:

3 0
3 years ago
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