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d1i1m1o1n [39]
3 years ago
15

a 10kg flowerpot is suspended from the end of a horizontal strut by a cable attached at 30° above the horizontal. If the strut h

as no mass what is tension in the cable

Physics
1 answer:
nikdorinn [45]3 years ago
8 0



T = tension force in the cable

m = mass of the flowerpot = 10 kg

θ = angle cable makes with the horizontal = 30 deg

there are horizontal and vertical components of tension force . The vertical component of tension force in cable is in opposite direction of the weight of the flowerpot. hence vertical component of tension balances the weight of flowerpot. so we can write

T Sinθ = mg

inserting the above value

T Sin30 = 10 x 9.8

T(0.5) = 98

T = 98/0.5

T = 196 N

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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
A car has a force of 0.27 N and an acceleration of 3m/s2 What is the car's mass?​
lyudmila [28]

Answer:

0.09 kg

Explanation:

f=0.27

a=3

m=f/a

m=0.27/3

m=0.09

7 0
3 years ago
A badger is running at a speed of 1 m/s. If the badger moves that was for 2600 seconds, how far will the badger travel?
katen-ka-za [31]

Answer:

It would be 2600

Explanation:

M/S stands for meters per second. If it moved 1 meter for 2600 seconds, than it would be 2600. You just multiply 2600 by 1! I hope this helps :D

8 0
3 years ago
The maximum pressure most organisms can survive is about 1000 times the atmospheric pressure. Only small, simple organisms such
Schach [20]

Answer:

h = 10000 m

Explanation:

The pressure applied at a depth of the liquid is given by:

P =ρgh

where,

P = Maximum Pressure to Survive = (1000)(Atmospheric Pressure)

P = (1000)(101325 Pa) = 1.01 x 10⁸ Pa

ρ = Density of sea water = 1025 kg/m³

g = 9.8 m/s²

h = maximum depth to survive = ?

Therefore,

1.01 x 10⁸ Pa = (1025 kg/m³)(9.8 m/s²)h

h = (1.01 x 10⁸ Pa)/(1025 kg/m³)(9.8 m/s²)

<u>h = 10000 m</u>

6 0
3 years ago
A vertical spring withk= 245N/m oscillates with an amplitude of 19.2cm when 0.457kg hangs from it. The mass posses through the e
Makovka662 [10]

Answer:

 y = -19.2 sin (23.15t) cm

Explanation:

The spring mass system is an oscillatory movement that is described by the equation

      y = yo cos (wt + φ)

Let's look for the terms of this equation the amplitude I

     y₀ = 19.2 cm

Angular velocity is

     w = √ (k / m)

     w = √ (245 / 0.457

     w = 23.15 rad / s

The φ phase is determined for the initial condition   t = 0 s ,  the velocity is negative v (0) = -vo

The speed of the equation is obtained by the derivative with respect to time

     v = dy / dt

     v = - y₀ w sin (wt + φ)

For t = 0

     -vo = -yo w sin φ

The angular and linear velocity are related v = w r

      v₀ = w r₀

      v₀ = v₀ sinφ

      sinφ = 1

      φ = sin⁻¹ 1

      φ = π / 4    rad

Let's build the equation

      y = 19.2 cos (23.15 t + π/ 4)

Let's use the trigonometric ratio π/ 4 = 90º

      Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a

       y = -19.2 sin (23.15t) cm

8 0
3 years ago
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