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PtichkaEL [24]
3 years ago
5

A group of students were investigating the force of gravity. They began by dropping a foam ball from a height of 3 meters into a

bucket of sand. The ball hit the sand in 0.306 seconds. They dropped additional balls of approximately the same diameter, but of different masses. Here is the data they collected. Based on this experiment and the collected data, what would their conclusion be?
A) Gravitational attraction is dependent on the mass of an object.
B) The greater the mass of the object the greater the acceleration due to gravity and the faster it will fall.
C) Mass does not affect the speed of falling objects assuming there is only the force of gravity acting on the objects.
D) Mass becomes a factor affecting the speed of a falling object when the object is dropped a great distance from the surface of the Earth.
Physics
1 answer:
kondaur [170]3 years ago
4 0

Answer:B

Explanation:

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An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached
Vera_Pavlovna [14]

Answer:

A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

B) Magnitude of charges = 1.206 × 10^(-6) C

Explanation:

We are given;

Spring constant;k = 350 N/m

Spring length;L = 0.39 m

Stretched length of spring;x = 0.022 m

A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

B) Force (F) required to stretch spring is given by the formula;

F = kx

Thus:

F = (350 × 0.022)

F = 7.7 N

Now, if we assume point charges, then the distance (r) between them will be given as:

r = (0.39 + 0.022) = 0.412 m

Coulomb's Law has a formula:

F = k(q1×q2)/r²

where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²

Making q1 × q2 the subject, we have;

(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

(q1 × q2) = 14.54 × 10^(-11) C

We are told that both charges are equal, thus; |q1| = |q2|

So;

q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

6 0
3 years ago
What radiation do remote controls use?
dalvyx [7]

Answer:

infra red radiation is used

3 0
3 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

The wavelength instead is just the distance between two consecutive crests, so

\lambda=4.8 m

And the wave speed is given by:

v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0
3 years ago
You are an astronaut in space far away from any gravitational field, and you throw a rock as hard as you can. The rock will:
Nesterboy [21]

Answer:

the rock will continue at the same speed unless it is affected by another force such as gravity and so if you threw it it will continue to move unless affected by a force

Explanation:

this is because Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

7 0
3 years ago
Consider a double slit experiment in the air. The wavelength of light is 500nm light, the screen is 1m away and two adjacent bri
Bess [88]

Answer: 0.75\ cm

Explanation:

Given

Wavelength of light \lambda=500\ nm

Screen is D=1\ m away

Distance between two adjacent bright fringe is \Delta y=\dfrac{\lambda D}{d}

When same experiment done in water, wavelength reduce to \dfrac{\lambda }{\mu}

So, the distance between the two adjacent bright fringe is \Delta y'=\dfrac{\lambda D}{\mu d}

Keeping other factor same, distance becomes

\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm

3 0
3 years ago
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