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Neporo4naja [7]
2 years ago
9

A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees to the horizontal. If the effective force

pulling the toy along the horizontal surface is 5N, calculate the tension in the rope

Physics
2 answers:
lana66690 [7]2 years ago
6 0
I'm thinking that you're supposed to divide. So you would divide 5 into 60 and get 12
myrzilka [38]2 years ago
6 0

As per the question the rope is inclined at an angle of 60 degree with the horizontal.

Let the tension produced in the rope is T.

The effective force which pulls the toy along horizontal direction is 5 N.

Resolving T into two components we get-

[1]\ \ Tsin\theta  in vertical upward direction.

[2]\ \ Tcos\theta in horizontal direction.

The vertical component is balanced by the weight of the body i.e

                            Tsin\theta=mg

Hence the toy moves due to the horizontal component.

                   Hence\ Tcos\theta=5 N

                         Tcos60= 5N

                          T= \frac{5}{cos60}

                                   = 5*2 N   [  cos60 = \frac{1}{2} ]

                                   = 10 N

Hence the tension produced in the rope is 10 N.

                         

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A shot-putter exerts a force of 0.142 kN on a shot, accelerating it to 22.75 m/s2. What is the mass of the shot?
Svetach [21]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Force and Power.

Since, according to the Newton's law,

Force = mass * Acceleration.

hence, here

Force = 142 N, accelration = 22.75 m/s2

hence, mass = 142/22.75

===> Mass = 6.24 Kg

hence the mass of the shot is 6.24 Kg

8 0
2 years ago
Last year a baseball player made 63 errors. This year he made 42. What percent decrease was there in the number of errors commit
Crank

Answer:

33.33 %

Explanation:

given,

error last year = 63

error this year = 42

percent decrease in the error = ?

to find the percentage difference in the error formula used is

   = \dfrac{difference}{original}\times 100

   = \dfrac{63-42}{63}\times 100

   = \dfrac{21}{63}\times 100

   = 33.33 %

Percentage decrease in the number of error is equal to 33.33%.

7 0
2 years ago
An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke
Elena-2011 [213]

Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

Explanation:

Total force required = Mass x Acceleration,

F = ma

Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

We need to accelerate the group of rocks from the road at 0.250 m/s²

That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

8 0
2 years ago
Determine the torque applied to the shaft of a car that transmits 225 hp
Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu

8 0
2 years ago
A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s
topjm [15]
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0
2 years ago
Read 2 more answers
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