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3 years ago

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An upward force is applied to a 6.0–kilogram box. This force displaces the box upward by 10.00 meters. What is the work done by

the force on the box?
a. 6.0 × 101 joules
b. -6.0 × 101 joules
c. 5.9 × 102 joules
d. -5.9 × 102 joules

1 answer:

devlian [24]3 years ago

3 0

Answer:

W ≈ 5.9 × $10^{2}$ J

Hence, option (c) is correct.

Given:

mass of box = 6 kg

Displacement of box = 10 m

To find:

Work done by the force = ?

Formula used:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

Solution:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

W = 6 × 9.8 ×10

W = 588 Joule

Thus, W ≈ 5.9 × $10^{2}$ J

Hence, option (c) is correct.

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2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos

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Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh$

there, no vertical movement

So, on neglect of height term

$P_{1}=P+\dfrac{1}{2}\rho v_{1}^2$

Where, P= atmospheric pressure

$\rho$ = air density

v = speed

Put the value in the equation

$P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)$

$P_{1}=2126.0\ lb/ft^2$

$P_{1}=\dfrac{2126.0}{144}$

$P_{1}= 14.76\ psi$

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{2}=P+\dfrac{1}{2}\rho v_{2}^2$

Put the value in the equation

$P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)$

$P_{2}=2246.17\ lb/ft^2$

$P_{2}=\dfrac{2246.17}{144}$

$P_{2}= 15.59\ psi$

(c). Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{3}=P+\dfrac{1}{2}\rho v_{3}^2$

Put the value in the equation

$P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)$

$P_{3}=2257.93\ lb/ft^2$

$P_{3}=\dfrac{2257.93}{144}$

$P_{3}= 15.68\ psi$

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

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A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha

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The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

$k=\dfrac{4\pi ^2m}{T^2}$

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

$k=\dfrac{4 \pi \times 0.2}{(1.2)^2}$

$k=\dfrac{39.48\times 0.2}{1.44}$

$k=\dfrac{7.90}{1.44}$

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

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1 year ago

8. Which explains how a plane mirror works?

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Answer:

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2 years ago

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig

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Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

the upper and the lower channels are at the same pressure (the atmospheric pressure).
the velocity of water in the upper channel is zero (v₁ = 0 m/s).
y₁ = 2.00 m (height of the upper channel)
y₂ = 0.00 m (height of the lower channel)
g = 9.81 m/s²
ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂

⇒ A = (350 m³/s)/(6.264 m/s)

⇒ A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4 ⇒ D = 2*√(A/π)

⇒ D = 2*√(55.873 m²/π)

⇒ D = 8.434 m ≈ 8.45 m

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L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

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