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2 years ago

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A municipal water supply is provided by a tall water tower. Water from this tower flows to a building. How does the water flow o

ut of a faucet on the ground floor of a building compare with the water flow out of an identical faucet on the second floor of the building

1 answer:

stealth61 [152]2 years ago

4 0

Answer:

THE ANSWER IS: Water flows more rapidly out of the ground-floor faucet.

Explanation:

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A 22-turn circular coil of radius 3.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plan

zloy xaker [14]

Answer:

23.5 mV

Explanation:

number of turn coil 'N' =22

radius 'r' =3.00 cm=> 0.03m

resistance = 1.00 Ω

B= 0.0100t + 0.0400t²

Time 't'= 4.60s

Note that Area'A' = πr²

The magnitude of induced EMF is given by,

lƩl =ΔφB/Δt = N (dB/dt)A

=N[d/dt (0.0100t + 0.0400 t²)A

=22(0.0100 + 0.0800(4.60))[π(0.03)²]

=0.0235

=23.5 mV

Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV

8 0

2 years ago

Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of

Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

$A=2\times10^{-6}m^2$

* The young modulus of the steel is,

$Y=2.1\times10^{11}\text{ Pa}$

Solution:

The young modulus of the steel in terms of the force and extension is,

$Y=\frac{F\times l}{A\times dl}$

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

$\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}$

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0

8 months ago

TP-4 When are lights required to be installed on a trailer?

kari74 [83]

Answer:

Light are required on every trailer

Explanation:

6 0

3 years ago

A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the m

Akimi4 [234]

Answer:

8.6 m

Explanation:

The motion of a soccer ball is a motion of a projectile, with a uniform motion along the horizontal (x-) direction and an accelerated motion along the vertical (y-) direction, with constant acceleration $a=g=-9.8 m/s^2$ towards the ground (we take upward as positive direction, so acceleration is negative).

The initial velocity along the vertical direction is

$v_{y0} = v_0 sin \theta = (26 m/s)(sin 30^{\circ})=13 m/s$

Now we can consider the motion along the vertical direction only. the vertical velocity at time t is given by:

$v_y(t)=v_{y0} +at$

At the point of maximum height, $v_y(t)=0$ , so we can find the time t at which the ball reaches the maximum height:

$0=v_{y0}+at\\t=-\frac{v_{y0}}{a}=-\frac{13 m/s}{-9.8 m/s^2}=1.33 s$

And now we can use the equation of motion along the y-axis to find the vertical position of the ball at t=1.33 s, which corresponds to the maximum height of the ball:

$y(t)=v_{y0}t + \frac{1}{2}at^2=(13 m/s)(1.33 s)+\frac{1}{2}(-9.8 m/s^2)(1.33 s)^2=8.6 m$

4 0

2 years ago

Does anyone know question 6?

vfiekz [6]

Answer:

Yes, the fuse will blow.

Explanation:

We'll begin by converting 50000mA to amperes (A) .this is illustrated below:

1000mA = 1A

Therefore, 50000mA = 50000/1000 = 50A.

From the question given above, the plug can only accommodate 5A. Now if 50000mA i.e 50A is passed through the plug, the plug will blow because the 50A is higher than what the plug can accommodate.

3 0

3 years ago