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2 years ago

Water at the top of Horseshoe Falls (part of Niagara Falls)

Physics

1 answer:

ale4655 [162]2 years ago

8 0

Answer:

$\theta= (-74.42)^{\circ} C$

Explanation:

Horizontal speed of water, $v_{xf}=9\ m/s$

Height, h = -53 (below pool)

We can find firstly the final vertical speed of the water using third equation of kinematics. So

$v^2_{yf}=u^2_{yi}+2(-g)h\\\\v^2_{yf}=2\times -9.8\times -53\\\\v_{yf}=32.23\ m/s$

Let $\theta$ is the angle where the falling water moving as it enters the pool. So,

$\tan\theta=\dfrac{v_{yf}}{v_{xf}}\\\\=\dfrac{-32.3}{9}\\\\=-74.42^{\circ} C$

Hence, the angle is (-74.42)°C.

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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to

Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

$p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa$

On the lower part of the hatch, there is a pressure equal to

$p_{bot}=p_{atm}=1.013\cdot 10^5 Pa$

So, the net pressure acting on the hatch is

$p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa$

which acts from above.

The area of the hatch is given by:

$A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2$

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

$F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N$

8 0

2 years ago

At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball

dedylja [7]

Answer:

Magnitude of net force will be 432.758 N

Explanation:

We have given x component of acceleration $a_x=760m/sec^2$

And vertical component of acceleration $a_y=770m/sec^2$

Mass of the ball m = 0.40 kg

So net acceleration $a=\sqrt{a_x^2+a_y^2}=\sqrt{760^2+770^2}=1081.896m/sec^2$

Now according to second law of motion

Force = mass × acceleration

So F = 0.40×1081.896 = 432.758 N

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2 years ago

How long does it take a glass of salt water to freeze than a glass of fresh water

max2010maxim [7]

A glass of salt water will take a slightly longer time & slightly lower temperature (28 F as compared to 32 F for fresh water) to freeze than a glass of fresh water.

Hope this helps!

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3 years ago

Which of the following statements best summarizes the main points of the passage?

Deffense [45]

Answer:the summery is ...i really don't know the picture is blurry and I cant see can you make it clear?

Explanation:

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1 year ago

You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di

kherson [118]

Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

116 km/1 h = 800 km/? h
800 = 116 · ?
? = 800/116
? = 6.89655172

I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

12 h - 6.89655172 h = 5.10344828 h

Using this remaining time and my remaining distance, I can calculate my average speed.

? km/1 hr = 400 km/5.10344828 h
5.10344828 · ? = 400
? = 400/5.10344828
? = 78.3783783148

My average speed during the last third of the distance is around 78 km/h.

8 0

2 years ago