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lisov135 [29]
1 year ago
12

Water at the top of Horseshoe Falls (part of Niagara Falls) is moving horizontally at 9.2 m/s as it goes off the edge and plunge

s 53 m to the pool below.
Part A
If you ignore air resistance, at what angle with respect to the vertical is the falling water moving as it enters the pool?
Physics
1 answer:
dangina [55]1 year ago
4 0

Water at the top of Horseshoe Falls (part of Niagara Falls) is moving horizontally at 9.2 m/s, the angle with respect to the vertical  is mathematically given as

\theta=15.93

<h3>What angle with respect to the vertical is the falling water moving as it enters the pool?</h3>

Generally, the equation for velocity is mathematically given as

V^2=u^2+2as

Therefore

V^{2}=0+2(9.81)(53)

V=32.2m/s

In conclusion, The angle

\theta =tan^{-1}\frac{(9.20m/s)}{(32.23m/s)}

\theta=15.93

Read more about Speed

brainly.com/question/4931057

#SPJ1

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The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?
pishuonlain [190]
The average dissipated power in a resistor in a ac circuit is:
P=I_{rms}^2 R
where R is the resistance, and I_{rms} is the root mean square current, defined as
I_{rms} =  \frac{I_0}{\sqrt{2}}
where I_0 is the peak value of the current. Substituting the second formula into the first one, we find
P=( \frac{I_0}{\sqrt{2} } )^2 R =  \frac{1}{2} I_0^2 R
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
I_0 =  \sqrt{ \frac{2 P}{R} }=  \sqrt{ \frac{2 \cdot 2.0 W}{47 \Omega} }=0.29 A
4 0
3 years ago
An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4x106m/s2. What is the ma
Jet001 [13]

Answer:

Explanation:

Force on electron in an electric field E = eE where E is electric field .

acceleration = eE / m where m is mass of electron .

Putting the values

4 x 10⁶ = 1.6 x 10⁻¹⁹ x E / 9.1 x 10⁻³¹

E = 22.75 x 10⁻⁶ N/C

The direction of electric field will be towards west ( opposite to east )

because of negative charge on electron .

7 0
3 years ago
Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu
Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes.  This is where density D appears as a  physical characteristic property of matter that establishes a relationship between the mass m of a body or substance and the volume V it occupies:

D=\frac{m}{V} (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is m=0.155g and th volume V=0.000235L=0.235mL

Solving (1) with these values:

D=\frac{0.155g}{0.235mL} (2)

D=0.660g/mL (3)

<h2>Answer B:</h2>

In this case the mass of a sample is m=4.71g and its density is D=3.63g/mL.

Isolating V from (1):

V=\frac{m}{D} (4)

V=\frac{4.71g}{3.63g/mL} (5)

V=1.297mL (5)

<h2>Answer C:</h2>

In this case the volume of a sample is V=0.293mL and its density is D=0.930g/mL.

Isolating m from (1):

m=D.V (6)

m=(0.930g/mL)(0.293mL) (7)

m=0.272g (8)

4 0
3 years ago
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A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

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6 0
2 years ago
If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?
lawyer [7]

Answer:

{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance =  \frac{50}{ {4.5}^{2} } }} \\  \\ { \rm{resistance = 2.5 \:  ohms}}

8 0
2 years ago
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