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algol13
3 years ago
14

A worker with spikes on his shoes pulls on rope that is attached to a box that is resting on a flat, frictionless frozen lake. T

he box has mass m, and the worker pulls with a constant tension T at an angle θ = 40 ∘ above the horizontal. There is a strong headwind on the lake, which produces a horizontal force Fw that is pointed in the opposite direction than the box is being pulled. Draw a free-body diagram for this system. Assume that the worker pulls the box to the right. If the wind force has a magnitude of 30 N, with what tension must the worker pull in order to move the box at a constant velocity?

Physics
1 answer:
earnstyle [38]3 years ago
5 0

Answer:

a

The free body diagram is shown on the first uploaded image

b

The tension on the rope is  T=39.16 \ N  

Explanation:

From the  question we are told that

    The mass of the box is  m

    The tension on the box is  T

     The angle at which it is pulled is  \theta =  40^o

     The force produced by the strong head wind is Fw =  30 \ N

At equilibrium the net force acting on the block along the horizontal axis is zero i.e

     Tcos \theta -F_w = 0

substituting values

     Tcos (40) -30 = 0

     Tcos (40)  = 30

     T(0.76604))  = 30

     T=39.16 \ N      

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eimsori [14]

Answer:

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7 0
3 years ago
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You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and
Elan Coil [88]

Answer: a) 10.4km

b) 0.62h

C) 17km/h

d) 9.1km/h

Explanation:

A) suppose you move in the positive direction of an x axis, from a first position of x1 = 0 to a second position of x2 at the station. That second position must be at x2 = 8.4 km + 2.0 km = 10.4 km.

B) time interval Dtwlk (= 0.50 h), but we don't know the driving time interval Dt/dr. However, we know that for the drive the displacement Dx/dr is 8.4 km and the average velocity vavg,dr is 70 km/h.

This average velocity is the ratio of the displacement for the drive to the time interval for the drive:

dv = dx/ dt

dt = dx/dv = 8.4/70 = 0.12h

t = ti + two

t = 0.12 + 0.50 = 0.62h

C) avg speed for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for the entire trip. 

Avg v = distance/time

10.4/0.62 = 16.8km/h

D) average speed is the ratio of the total distance you move to the total time interval you take to make that move. The total distance is 8.4 km + 2.0 km + 2.0 km = 12.4 km. The total time interval is 0.12 h + 0.50 h + 0.75 h = 1.37 h.

Avg speed = 12.4/1.37 = 9.1km/h

6 0
4 years ago
A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

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4 years ago
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Fofino [41]

Two copper spheres are currently 1.2 meters apart. One sphere has a charge of +2.2•10-4 C and the other has a charge of -8.9•10-4 C, then the the force between the charged spheres is the force attractive.

<h3>What is Coulomb's law?</h3>

Coulomb's law can be expressed as that the electrical force between two charged bodies is directly proportional to the product of the quantity of charge on the bodies and inversely proportional to the square of the separation distance between the two bodies.

As given in the problem two copper spheres are currently 1.2 meters apart.One sphere has a charge of +2.2•10-4 C and the other has a charge of - 8.9•10-4 C.

As per coulomb's law, opposite charges attract each other.

Thus, If one sphere has a charge of +2.2•10-4 C and the other has a charge of -8.9•10-4 C, then the the force between the charged spheres is the force attractive.

Learn more about Coulomb's law here ;

brainly.com/question/506926

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5 0
1 year ago
The energy released during a nuclear reaction is a result of
astra-53 [7]

Answer:

the conversion of matter into energy

Explanation:

A nuclear reaction is a reaction in which there is a change in the nuclei of the atoms involved. The two most important examples of nuclear reactions are:

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In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei: this means that part of the mass of the initial nuclei has been converted into energy, according to Einstein's formula:

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