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evablogger [386]

3 years ago

6

A tow truck drags a stalled car along a road. The chain makes an angle of 30° with the road and the tension in the chain is 1000

N. How much work is done by the truck in pulling the car 1 km? (Round your answer to the nearest ten thousand.) joules

1 answer:

statuscvo [17]3 years ago

6 0

To develop this problem it is necessary to apply the work theorem. This tells us that the energy on a body is equivalent to the product between the Force applied on it and the distance traveled. The vector product of these measures can be expressed as

$W = \vec{F}\times \vec{x}$

And its magnitude is equivalent to

$W = Fx cos\theta$

Where $\theta$ is the angle formed between the applied force and the line of displacement of the body.

Our values are given as,

$\theta = 30\°$

$F= 1500N$

$x = 1000m$

Applying this value and replacing we have,

$W = Fxcos\theta$

$W = (1500)(1000)cos(30)$

$W = 1.299*10^6J$

Therefore the work done by the truck is 1.299 Mega Jules

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A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735

Irina18 [472]

Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.

7 0

3 years ago

Read 2 more answers

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

2. A hanging wind-chime on a calm day would have kinetic or potential energy?

rjkz [21]

Answer:

it would have potential energy

7 0

3 years ago

Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential

mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0

3 years ago

Two horses are pulling a box in two different directions as shown in the below image. The image shows one 30.0 N force due north

Novay_Z [31]

Explanation:

'What is the magnitude of the force needed to stop the horses and bring the box into equilibrium?' ≈42N; according to the vectors rules.

'Where would you locate the rope to apply the force?' - in point D.

PS. zoom out the attached picture.

4 0

3 years ago