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Effectus [21]
3 years ago
12

A 3.0kg mass tied to a string

Physics
1 answer:
dem82 [27]3 years ago
5 0

Answer:

\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force (\sf F_c).

\boxed{ \bold{ T = F_c  =  \frac{m {v}^{2} }{r} }}

Substituting value of m, v & r in the equation:

\sf \implies T =  \frac{3 \times  {20}^{2} }{0.4}  \\  \\  \sf \implies T = \frac{3 \times 400}{0.4}  \\  \\  \sf \implies T =3 \times 1000 \\  \\  \sf \implies T =3000 \: N \\  \\ \sf \implies T =3 \: kN

\therefore

Tension in the string (T) = 3 kN

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How high can a 40 N force move a load, when 395 J of work is done?
Juliette [100K]

Answer:

9.875

Explanation:

w=f×s

395=40×s

make s the subject of the formula

s=395/40

=9.875

7 0
3 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti
ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = 6.31\times 10^{4} T

Solution:

To find the number of turns, N, we use the given formula:

B = \frac{\mu_{o}NI}{2R}

Therefore,

N = \frac{2BR}{\mu_{o}I}

N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}

N = 22.74 = 23 turns (approx)

 

8 0
2 years ago
How do you find the direction of displacement?
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It is a distance force
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