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3 years ago

A machine has an efficiency of 80%. How much work must be done on the machine so to make it do 50,000 J of output work?

Physics

1 answer:

san4es73 [151]3 years ago

6 0

Find 10 % of 50,000 = 5000
double it to make 10000
add on to 50000 to make 60000
60000 is your answer

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Please write 4-5 sentences, please!! How would you describe the climate of Colorado? Find 1 piece of evidence from online to def

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Answer:

The climate in colorado is combination of high elevation, midlatitude, and continental interior geography results in a cool, dry, and invigorating climate. The average annual temperature for the state is 43.5 degrees Fahrenheit (F), which is 13.7 degrees below the global mean

Explanation:

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3 years ago

When you are aware that you are having a dream and can control what you do

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Explanation:

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3 years ago

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A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

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3 years ago

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

slega [8]

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

8 0

3 years ago

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb

Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 $in^{2}$

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = $\frac{F}{A}$

⇒ P = $\frac{6.4}{0.4}$

⇒ P = 16 $\frac{lb}{in^{2} }$

This is the pressure inside the cylinder.

Let force applied on the brake pad = $F_{1}$

Area of the brake pad ( $A_{1}$ )= 1.7 $in^{2}$

Thus the pressure on the brake pad ( $P_{1}$ ) = $\frac{F_{1} }{A_{1} }$

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = $P_{1}$

⇒ 16 = $\frac{F_{1} }{A_{1} }$

⇒ $F_{1}$ = 16 × $A_{1}$

Put the value of $A_{1}$ we get

⇒ $F_{1}$ = 16 × 1.7

⇒ $F_{1}$ = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = $\frac{F_{1} }{4}$ = $\frac{27.2}{4}$ = 6.8 lb

⇒ The force applied on one wheel during braking.

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3 years ago