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3 years ago

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

Physics

1 answer:

ale4655 [162]3 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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Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r

KIM [24]

The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
$F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N$

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3 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

4 years ago

Read 2 more answers

Is a magnetic reversal a threat to life on Earth ? Whay do you say?

KATRIN_1 [288]

I would say yes, a magnetic reversal a threat to life on Earth. Base on articles, it is the end of the world if that happens. <span>Some people believe global cataclysm will occur when Earth's magnetic poles reverse. When north goes south, they say, the continents will lurch in one direction or the other, triggering massive earthquakes, rapid climate change and species extinctions.</span>

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3 years ago

A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.

blondinia [14]

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

$F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N$