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3 years ago

A pendulum on a grandfather clock

Physics

1 answer:

egoroff_w [7]3 years ago

5 0

4.0 ilynits per second Alaskan es muy du facial in the oscillates 1.99

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With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t

VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

$v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N$

Therefore, the tension of the rope is 34.95 N

4 0

3 years ago

Why is it cold in the mountains if heat rises?

hjlf

Answer:

look at explanation

Explanation:

If heat rises, then why is it so cold at the top of a mountain? Heat does indeed rise. More specifically, a mass of air that is warmer than the air around it expands, becomes less dense, and will therefore float atop the cooler air. ... So when warm air rises, it cools off.

6 0

2 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago

Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m

const2013 [10]

Answer:

The answer to your question is vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

hmax = vo²sinФ/ 2g

Solve for vo²

vo² = 2ghmax / sinФ

Substitution

vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

vo² = 19.62(1.3) / 0.866

vo² = 25.51 / 0.866

vo² = 29.45

Result

vo = 5.43 m/s

5 0

3 years ago

Help immediately! Answer I need help.

garri49 [273]

Answer:

0.80 m

Explanation:

elastic potential energy formula

elastic potential energy = 0.5 × spring constant × (extension) 2

4 0

2 years ago