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Fiesta28 [93]
2 years ago
14

Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an

angle θ with the direction of motion of the horses and the barge. Each horse is traveling at a constant speed v.
a. How much work W is done by each horse in a time t? Express the work in terms of the quantities given.
b. How much power P does each horse provide? Express your answer in terms of the quantities given.
Physics
1 answer:
morpeh [17]2 years ago
8 0

Answer:

<em>a) Fvt cosθ</em>

<em>b) Fv cosθ</em>

<em></em>

Explanation:

Each horse exerts a force = F

the rope is inclined at an angle = θ

speed of each horse = v

a) In time t, the distance traveled d = speed x time

i.e d = v x t = vt

also, the resultant force = F cosθ

Work done W = force x distance

W = F cosθ x vt = <em>Fvt cosθ</em>

<em></em>

b) Power provided by the horse P = force x speed

P = F cosθ x v

P = <em>Fv cosθ</em>

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Further Explanation:

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Subject: physics

Level: High School

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5 0
3 years ago
Read 2 more answers
If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100
SCORPION-xisa [38]

Answer:

Charge enter a 0.100 mm length of the axon is 8.98\times 10^{-12} C

Explanation:

Electric field E at a point due to a point charge is given by

E=k \frac{q}{r^2}

where k is the constant =9.0 \times 10^9  Nm^2 / C^2

q is the magnitude of point charge and r is the distance from the point charge

Charges entering one meter of axon is 5.\times 10^{11} \times (+e)

Charges entering 0.100 mm of axon is 5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}

substituting the value of +e=1.6\times 10^{-19} C in above equation, we get charge enter a 0.100 mm length of the axon is

q=5.\times 10^{11} \times1.6\times 10^{-19}  \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C

3 0
2 years ago
For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a
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8 0
2 years ago
7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle
jasenka [17]

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

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Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
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To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

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2 years ago
The force sensor measures the force on the sensor due to the bumper, but the cart's momentum change arises from the force on the
Irina-Kira [14]

Answer:

Answered

Explanation:

1 and 3 are necessary

Every bit of force applied to the bumper will be transmitted to the cart EXCEPT for the force needed to accelerate the bumper. This is the net force on the bumper.

If the bumper was heavy then a significant amount of force might be needed to accelerate the bumper so the amount transmitted to the cart would be substantially reduced.

If the net force on the bumper is small then the amount transmitted to the cart is almost the entire force applied.

5 0
2 years ago
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