Is the answer system

You are performing an experiment that requires the highest possible energy density in the interior of a very long solenoid. Whic

Alinara [238K]

Answer:

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

Explanation:

As we know that energy density depends on the strength of the magnetic field. The magnetic field strength depends on the no of turns of the solenoid and the current passing through it. The greater the number of turns per unit length, greater the current passing through it, more stronger the magnetic field is. As

B = μ₀nI

n = no of turns

I = current through the wire

So the right options are

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

5 0

3 years ago

What is the marble's range if it is fired horizontally from 1.6 m above the ground?

cupoosta [38]

Answer:

The answer is 6.40 meters.

Explanation:

The speed v = √(2gh)

v = √( 2×9.8×6.4) = 11.2 m/s

After, finding the time it takes to hit the ground from a height of 1.6 meters.

time = √(2H÷g)

time = √(2×1.6÷9.8)

time = 0.5714 seconds.

Horizontal distance is speed × time = 11.2 × 0.5714 = 6.40 meters.

5 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

Time taken by the rocket to fall from maximum height is 20.29 seconds

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago

A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear

JulijaS [17]

Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval (−L,0] for some large number L. The voltage V as a function of x on the interval (0,∞) is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element dx′, we have λ=dqdx′.V(x)=1/4πϵ0∫linedq/r=λ/4πϵ0∫−L0dx/x−x′=λ/4πϵ0(ln|x+L|−ln|x|)

5 0

3 years ago

Please Its urgent I need your HELP!!!!!!!!!!!!!!!!!!!!!!!

ozzi

Raising the temperature results in the radiator giving off photons of high-energy ultraviolet light. As heat is added, the radiator emits photons across a wide range of visible-light frequencies

4 0

2 years ago