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stealth61 [152]
3 years ago
12

A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak

es an angle of 40 degrees. What is the tension (in Newtons) on the right rope?
Physics
1 answer:
GarryVolchara [31]3 years ago
7 0

Explanation:

It is given that,

Mass of the box, m = 100 kg          

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.  

From the attached figure, the x and y component of forces is given by :

T_{1x} =-T_1 cos (20)

T_{2x} = T_2 cos (40)

mg_x = 0

T_{1y} = T_1 sin (20)

T_{2y} = T_2 sin (40)

mg_y= -mg

Let R_x and R_y is the resultant in x and y direction.

R_x=-T_1 cos (20)+T_2 cos (40)+0

R_y=T_1 sin(20)+T_2 sin(40)-mg

As the system is balanced the net force acting on it is 0. So,

-T_1 cos (20)+T_2 cos (40)+0=0.............(1)

T_1 sin(20)+T_2 sin(40)-100\times 9.8=0..................(2)

On solving equation (1) and (2) we get:  

T_1=866.86\ N (tension on the left rope)

T_2=1063.36\ N (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.                            

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A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is
andriy [413]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

4 0
3 years ago
Explain how streams can erode soil and transport materials?
vladimir1956 [14]
The eroded rock and soil materials that are transported downstream by a river are called its load. A river transports, or carries, its load in three different ways: in solution, in suspension, and in its bed load. Mineral matter that has been dissolved from bedrock is carried in solution. Common minerals carried in solution by rivers include dissolved calcium, magnesium, and bicarbonate. Most of a river’s solution load comes from groundwater seeping into the river. Before it reaches the stream,thegroundwaterhastraveledthroughfracturesinthebedrock, chemically eroding rock along the way. When river water looks muddy, it is carrying rock material in suspension. Suspended material includes clay, silt, and fine sand. Although these suspended materials are heavier than water, the turbulence of the stream flow stirs them up and keeps them from sinking. Turbulence includes swirls and eddies that form in water as a result of friction between the stream and its channel. The faster a stream flows, the more turbulent and muddy it becomes. A rough or irregular channel also increases turbulence. A river may also transport rock materials in its bed load. The bed load consists of sand, pebbles, and boulders that are too heavy to be carried in suspension. These heavier materials are moved along the streambed, especially during floods. Boulders and pebbles roll or slide along the river bed. Large sand grains are pushed along the bottom in a series of jumps and bounces. The relative amounts of a river’s load that are carried in solution, in suspension, and in the bed load depend on the nature of the river, the climate, the type of bedrock, and the season of the year. As a general rule, most of the load carried by the world’s streams and rivers is carried in suspension. The size of a river’s suspended load increases with human land use. Road and building construction and removal of vegetation make it easier for rain to wash sediment into streams and rivers.
8 0
3 years ago
A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from
Naddik [55]

Complete Question

The complete question iws shown on the first uploaded image  

Answer:

a

    y \to z \to x

b

  x \to z \to y

Explanation:

Now looking at the diagram let take that the magnetic field is moving in the x-axis

 Now the magnetic force is mathematically represented as

             F =  I L x B

Note (The x is showing cross product )

Note the force(y-axis) is perpendicular to the field direction (x-axis)

Now when the loop is swinging forward

 The motion of the loop is  from   y to z to to x to y

Now since the force is perpendicular to the motion(velocity) of the loop

Hence the force would be from z to y and back to z  

and from lenze law the induce current opposes the force so the direction will be from y to z to x

Now when the loop is swinging backward

   The motion of the induced current will now be   x to z to y

 

5 0
3 years ago
Can someone explain to how to calculate this
Karo-lina-s [1.5K]

answer

option d is the correct answer

explanation

as we know frequency is equal to 1 /t

f= 457 Hz

t=1

SO, 1/457

=0.0022sev

3 0
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(a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere’s velocity as a function of tim
jeka94

Answer:

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