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3 years ago

12

A pendulum bob is made from a sphere filled with water. What would happen to the frequency of vibration of this pendulum if the

sphere had a hole in it that allowed the water to leak out slowly?

1 answer:

Nady [450]3 years ago

7 0

Answer:

The frequency would not change because the frequency of a simple pendulum does not depend on mass.

Explanation:

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A 4-kg object is moving with a speed of 5 m/s at a height of 2 m. The kinetic

tatyana61 [14]

Hello!

$\large\boxed{KE = 50J}$

Use the formula for kinetic energy:

$KE = \frac{1}{2}mv^{2}$

Plug in the given mass and velocity:

$KE = \frac{1}{2} (4)5^{2}$

Simplify:

$KE = \frac{1}{2} (100)\\\\KE = 50 J$

7 0

3 years ago

La luz roja visible tiene una longitud de onda de 680 nanómetros (6,8 x 10-7 m). La velocidad de la luz es de 3.0 x108 m / s. ¿C

Lina20 [59]

Answer:

Frequency, $f=4.41\times 10^{14}\ Hz$

Explanation:

Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?

It is given that,

Wavelength of a visible red light is, $\lambda=6.8\times 10^{-7}\ m$

Speed of light is, $c=3\times 10^8\ m/s$

We need to find the frequency of visible red light. It can be calculated using below relation.

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz$

So, the frequency of visible red light is $4.41\times 10^{14}\ Hz$ .

3 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

8 a Name the bones that articulate (join together) in the knee joint

Marianna [84]

Answer:

The Femur and the Tibia

8 0

3 years ago

A 35-gram stainless steel ball on a track is moving at a velocity of 9 m/s. On the same track, a 75-gram stainless steel ball is

elena-s [515]

Answer:

4.2 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v

315 g m/s − 525 g m/s = -525 g m/s + (75 g) v

315 g m/s = (75 g) v

v = 4.2 m/s

5 0

3 years ago

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