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4 years ago

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A swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across a river that has a downstream curre

nt of 3.75 m/s. 46.
a. what must the swimmer’s direction be?
b. what is his velocity relative to the bank?

1 answer:

e-lub [12.9K]4 years ago

7 0

Send more info please what area are u on

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A man at point A directs his rowboat due north toward point B, straight across a river of width 100 m. The river current is due

prohojiy [21]

Answer:

1.35208 m/s

Explanation:

Speed of the boat = 0.75 m/s

Distance between the shores = 100 m

Time = Distance / Speed

$Time=\frac{100}{0.75}=133.33\ s$

Time taken by the boat to get across is 133.33 seconds

Point C is 150 m from B

Speed = Distance / Time

$Speed=\frac{150}{\frac{100}{0.75}}=1.125\ m/s$

Velocity of the water is 1.125 m/s

From Pythagoras theorem

$c=\sqrt{0.75^2+1.125^2}\\\Rightarrow c=1.35208\ m/s$

So, the man's velocity relative to the shore is 1.35208 m/s

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4 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.

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

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

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

By making the spring constant k the subject of the formula:

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

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

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

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

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3 years ago

When you drive through deep water, you should dry out your brakes by _____________. pumping your brakes and driving in low gear

Mandarinka [93]

Answer:

By Applying pressure to the brakes

Explanation:

Driving cars through deep water that is more than 10cm can make the cars to float. Most modern cars are usually water- tight so they can start to float through water that is about 30cm deep, fast moving water is very powerful so one needs to be very careful when driving.

If the brakes are wet test them by pressing or tapping on them gently.

You can as well dry brakes by driving in low gear and applying pressure to the brakes.

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3 years ago