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larisa [96]
3 years ago
8

A pendulum has a 0.35\ \text{kg}0.35 kg0, point, 35, space, start text, k, g, end text mass oscillating at a small angle from a

light string of length 1.0\ \text{m}1.0 m1, point, 0, space, start text, m, end text. What is the frequency of oscillation?
Physics
1 answer:
expeople1 [14]3 years ago
4 0

Answer:

The frequency of oscillation of the simple pendulum is 0.49 Hz.

Explanation:

Given that,

Mass of the simple pendulum, m = 0.35 kg

Length of the string to which it is attached, l = 1 m

We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{1}} \\\\f=0.49\ Hz

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.

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Derive an expression for the acceleration of the car. Express your answer in terms of D and vt Determine the time at which the s
Bezzdna [24]

Answer:

V(car) = V(truck) at t = Dt/2

acceleration = v(car) = D/t^2

Explanation:

acceleration = v(car) = D/t^2

Since the average velocities must be the same, the car's final velocity must be twice the trunk velocity assuming the car start with zero velocity, since acceleration remain the same throughout the journey velocities at half-time point must be equal.

3 0
4 years ago
Which of these has components that can have a varying ratio?
Travka [436]
I say C. But I’m not 100% sure so check it first
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3 years ago
What is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?
Art [367]

CORRECT ANSWER:

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

STEP-BY-STEP EXPLANATION:

The complete question from book is

According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.

c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.

d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.

e- None of the other answer options is correct.

3 0
4 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
azamat

Answer:

a. Speed = 1.6 m/s

b. Amplitude = 0.3 m

c. Speed = 1.6 m/s

Amplitude = 0.15 m

Explanation:

a.

The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:

frequency = \frac{1}{2(2\ s)} = 0.25 Hz

The wavelength of the wave is the distance between consecutive crests of wave. Therefore,

Wavelength = 6.4 m

Now, the speed of the wave is given as:

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

<u>Speed = 1.6 m/s</u>

<u></u>

b.

Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:

Amplitude = (0.5)(0.6 m)

<u>Amplitude = 0.3 m</u>

<u></u>

c.

frequency = \frac{1}{2(2\ s)} = 0.25 Hz

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

<u>Speed = 1.6 m/s</u>

<u></u>

Amplitude = (0.5)(0.3 m)

<u>Amplitude = 0.15 m</u>

8 0
3 years ago
What muscle movement do we use to close a joint.
marin [14]

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