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2 years ago

In which of these samples do the molecules mist likely have the least kinetic energy

Physics

1 answer:

KiRa [710]2 years ago

8 0

Provide the examples again, no samples are presents.

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A machine that operates a ride at the fair requires 2500 J to lift a 294 N child 5.0 m. What is the efficiency of this machine?

NeX [460]

Answer:

η = 58.8%

Explanation:

Work is defined as the force applied by the distance traveled by the body.

$W =F*d$

where:

W = work [J] (units of joules)

F = force = 294 [N]

d = distance = 5 [m]

$W = 294*5\\W = 1470 [J]\\$

Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

$efficiency = W_{done}/W_{required}\\efficiency = 1470/2500\\efficiency = 0.588 = 58.8%$

5 0

2 years ago

Jolene travels north 5 miles and then goes west 3 miles before coming straight back south 2 miles. What is her distance

Maurinko [17]

Answer:

mnbhngbfcvdxc

Explanation:

8 0

2 years ago

Hey!

pav-90 [236]

Answer:

Hydraulic pressure exerted on glass slab, ρ=10 atm

Bulk modulus of glass, B=37×10^9 Nm^−2

Bulk modulus, B=P/(ΔV/V)

where,

ΔV/V= Fractional change in volume

ΔV/V=P/B

=10×1.013×10^5 /(37×10 ^9)

=2.73×10^-5

Therefore, the fractional change in the volume of the glass slab is 2.73×10^-5

Hope it helps

3 0

2 years ago

Read 2 more answers

Suppose a photon with an energy of 1.60 eV strikes a piece of metal. If the electron that it hits loses 0.800 eV leaving the met

JulsSmile [24]

To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.

From the theory we could consider that the energy change is subject to

$\Delta E = E_0 -W_f$

Where

$E_0 =$ Initial Energy

$W_f =$ Energy loses

Replacing we have that

$\Delta E = 1.6-0.8$

$\Delta E = 0.8eV$

Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV

7 0

2 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.

7 0

2 years ago