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3 years ago

What is the reactive force when a fish swims through water?

1 answer:

7 0

The answer is D

This is an example of Newton's third law of motion.

The active force is the fish against the water, so the reactive force would be the reverse, the equal force of the water on the fish.

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If you were to walk at a constant speed 20m/s for 30 seconds, how far would you walk?

lana [24]

Answer:

600m

Explanation:

30×20 at a constant speed is 600m.

6 0

3 years ago

If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return?

iren [92.7K]

Same speed, because mass is neglected. The things that affect the speed are the distance and speed of the rock.

5 0

3 years ago

Weekend A Assignment Differentiate between forced and damped oscillation

4vir4ik [10]

Answer:

A damped oscillation means an oscillation that fades away with time while Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system.

Explanation:

Damping is the reduction in amplitude (energy loss from the system) due to overcomings of external forces like friction or air resistance and other resistive forces. ... When a body oscillates by being influenced by an external periodic force, it is called forced oscillation.

<h2>Hope this helped </h2>

Welcome 

6 0

3 years ago

When an object like a tree is illuminated by the sun, and you are looking toward the tree, light rays leave the object _________

Dima020 [189]

Objects absorb and reflect light differently depending on their physical characteristics, such as their shape or composition. Thanks to the reflection we can see the objects. Reflection can be defined as the change of direction of a wave, which, when in contact with the separation surface between two changing means, returns to the point where it originated. When the light illuminates the object, such as the tree, the rays of light will disperse in all directions allowing observation.

The correct answer is A. From every point on the surface of the tree, and in every direction

6 0

3 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$