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Korvikt [17]
3 years ago
15

Two men decide to use their cars to pull a truck stuck in mud. They attach ropes and one pulls with a force of 598 N at an angle

of 29◦ with respect to the direction in which the truck is headed, while the other car pulls with a force of 941 N at an angle of 23◦ with respect to the same direction. 598 N 29 ◦ 941 N 23 ◦ What is the net forward force exerted on the truck in the direction it is headed? Answer in units of N
Physics
1 answer:
kondaur [170]3 years ago
5 0

Answer:

magnitude of the force is

F = 1537 N

direction of the force is given as

\theta = 25.3 degree

Explanation:

As we know that the first force is 598 N at 29 degree

so we have

F_1 = 598(cos29 \hat i + sin29 \hat j)

F_1 = 523 \hat i + 290 \hat j

Now another force is 941 N at 23 degree

F_2 = 941(cos23\hat i + sin23\hat j)

F_2 = 866.2 \hat i + 367.7 \hat j

so we will have

F = F_1 + F_2

F = (523 + 866.2)\hat i + (290 + 367.7)\hat j

F = 1389.2\hat i + 657.7 \hat j

magnitude of the force is

F = \sqrt{1389.2^2 + 657.7^2}

F = 1537 N

direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{657.7}{1389.2}

\theta = 25.3 degree

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Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
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A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.
Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
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4 years ago
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If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le
GarryVolchara [31]

Answer:

  F_{y} = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

        B₀ - W = 0

        B₀ = W

        ρ_fluid g V_fluid = W

the volume of the fluid is the area of ​​the cube times the height it is submerged

      V_fluid = A y  

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

        B - W = m a

        ρ_fluid g A (y₀ + y) - W = m a

        ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

       ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

      ρ_fluid g A y = m a

      this equation the net force is

      F_{y} = ( ρ_fluid g A) y

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8 0
3 years ago
Describe each class of lever and explain to characteristics of each
Nataly [62]

-- Class I lever

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Example:  a see-saw

-- Class II lever

The load is between the fulcrum and the effort.

The Mechanical Advantage is always greater than 1 .

Example:  a nut-cracker, a garlic press

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Komok [63]

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity  F=W= mg   [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

                                        =132 kg ×9.8 m/s^2

                                        =1293.6 kg m/s^2

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As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

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Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

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From Newton's second law of motion we know that net force is the product of mass and acceleration i.e  

                                   F_{net} =ma  [Here a is the acceleration]

                                             a =\frac{F_{net} }{m}

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In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

 we know that F= ma

                           =m×0

                            =0 N

Hence they will not get any force when they will descend with a uniform speed.


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3 years ago
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