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wolverine [178]
3 years ago
6

A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold welghs 198 N and is 3.2 m long.

what is the tension in each rope whien the 6oo-N worker stands 1.12 m from one end? smaller tension arger tension
Physics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

T_2 = 309 N

T_1 = 489 N

Explanation:

As we know that total tension in both the ropes is counter balancing the weight of scaffold and worker both

so here we will have

T_1 + T_2 = (m + M)g

now we have

T_1 + T_2 = 198 + 600 = 798 N

now we also know that net torque due to both tension force in the string with respect to the position of worker must be zero so that platform will remain in equilibrium and horizontal in position

so here we will have

T_1(1.12) + (198)(1.6 - 1.12) = T_2(3.2 - 1.12)

T_1+ 84.86 = 1.86 T_2

now from above two equations we will have

(1.86 T_2 - 84.86) + T_2 = 798

T_2 = 309 N

also we have

T_1 = 489 N

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A ball is thrown straight upward and returns to the thrower’s hand after 1.8 s in the air. A second ball is thrown at an angle o
zysi [14]

Answer:

U = 9.1 m/s

Explanation:

from the question we are given the following

time (t) = 1.8 s

angle = 23 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below

v = u + 0.5at

  • The final velocity (v) is zero since the ball comes to rest
  • The time (t) it takes to get to the maximum height would  be half the time it is in the air, t = 0.5 x 1.8 = 0.9

therefore

0 = u - (0.5 x 9.8 x 0.9)

u = 7.9 m/s

for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore

U sin 60 = 7.9

U = 7.9 ÷ sin 60

U = 9.1 m/s

5 0
3 years ago
A book is on the table. If the weight of the book is 25 newtons, what is the magnitude and direction of the normal force?
Andrej [43]
I'm not sure about the magnitude, but the direction of the normal force is upward. 
4 0
3 years ago
Read 2 more answers
When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le
77julia77 [94]

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

x+2.93

Both cover a magnitude of 8.32 ft, therefore

x +(x+2.97) = 8.32

Now solving for x we have,

x + (x + 2.93) = 8.32

2x + 2.93 = 8.32

2x = 8.32 - 2.93

x = \frac{ 8.32 - 2.93}{2}

x = 2.695 ft

Therefore the shorter stick is 2.695ft long.

7 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of
Naya [18.7K]

Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t²    ..............1

and

depth = velocity of sound × time    .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1  and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

3 0
3 years ago
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