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3 years ago

6

while playing her guitar , karen plucks one string with increasin levels of force. what effect does this have on the sound produ

ced?

1 answer:

kirza4 [7]3 years ago

3 0

The sound will get louder.

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A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 6.5 cm. The

GREYUIT [131]

Explanation:

Since the chain isn't slipping, the pedal sprocket and wheel sprocket have the same linear velocity.

First, we find the angular velocity of the pedal sprocket:

1.1 rev/s × 2π rad/rev = 2.2π rad/s ≈ 6.9 rad/s

Next, we find the linear velocity of the pedal sprocket:

v = ωr = (6.9 rad/s)(9.5 cm) = 66 cm/s

Now we find the angular velocity of the wheel sprocket:

ω = v/r = (66 m/s) / (6.5 cm) = 10. rad/s

So the linear velocity of the bike is:

v = ωr = (10. rad/s) (0.65 m) = 6.6 m/s

The linear velocity of the bike is proportional to the angular velocity of the pedal. So if the bicyclist wanted to move at a different speed, say 5.5 m/s, we could find the new angular velocity of the pedal by writing a proportion:

1.1 rev/s / 6.6 m/s = x / 5.5 m/s

x = 0.92 rev/s

Or, 1/x = 1.1 seconds per revolution.

7 0

4 years ago

A vertical spring with spring constant 23.15 N/m is hanging from a ceiling. A small object is attached to the lower end of the s

umka2103 [35]

Answer:

m= 1.47 kg

Explanation:

Given:

spring constant, K = 23.15 N/m

Displacement, x= 19.79 cm = 0.1979 m

at, x₁= 7.417 cm, v₁= 0.7286 m/s

Now,

x = 19.79 cos ( ωt)

on substituting the values, we get

7.417 x 10⁻² = 19.79 x 10⁻² cos (ωt)

or

cos(ωt) = 0.374

or

ωt = 67.98°

also

v = -0.1979×ωsin (ωt)

also

$\omega=\frac{2\pi}{T}$

on substituting the values in the above equation, we get

0.7286 = -0.1979 $\frac{2\pi}{T}$ sin ( 67.98°)

3.68 =- $\frac{2\pi}{T}$ (0.927)

or

T = 1.582 sec

also,

$T=2\pi\sqrt{\frac{m}{k}}$

where, m is the mass

on substituting the values, we have

$1.582=2\pi\sqrt{\frac{m}{23.15}}$

on squaring both sides and solving, we have

m= 1.47 kg

8 0

3 years ago

"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and

spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

8 0

3 years ago

The final image from a simple two-lens telescope is:

s2008m [1.1K]

A telescope with only one lens is built to photograph with.

A telescope with two lenses is built to look through. It produces
an image for your eye to look at. The image is virtual and inverted.

4 0

3 years ago

The velocity of a proton emerging from a Van de Graaff accelerator is 25.0% of the speed of light. What was the equivalent volta

LenaWriter [7]

Explanation:

Below is an attachment containing the solution.

8 0

4 years ago