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Dmitry_Shevchenko [17]
3 years ago
12

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t

he point of ejection are 1 meter/second and 0.5 meters/second respectively, calculate the ratio of their horizontal distances of impact.
Physics
2 answers:
IrinaK [193]3 years ago
4 0
 t1 = √2h1/g = √2*0.5/9,8 = 0.319 sec 
t2 = √2h2/g = √2*1.0/9,8 = 0.451 sec 

In which t = times for the vertical movement
h = height
g= gravity (we use standardized measurement of 9.8)

d1 = 1*0.319 = 0.319 m 
d2 = 0.5*0.451 = 0.225 m 

in which d = Horizontal distance

ratio
= di : d2
= 0.319 : 0.225    

 = 3.19 : 2.25
Lubov Fominskaja [6]3 years ago
3 0

Answer:

1.41:1

Explanation:

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4 years ago
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow
Helen [10]

Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

6 0
3 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k
ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

g=\frac{GM}{R^{2}}

g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

v^{2}=u^{2}-2gh

0 = v² - 2 x 1.62 x 2.55 x 10^6

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v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0
3 years ago
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