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____ [38]
3 years ago
8

Joseph studied whether different materials can block certain electromagnetic waves by testing television reception in different

parts of the house. At each part of the house, Joseph used a different antenna. The experiment could have been improved by _______________________________. Question 10 options: Testing reception of a different channel during each trial. Using the same antenna during each trial. Measuring the antenna length in centimeters. Measuring the distance to the television station.
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0
Using the same antenna. Now he doesn't know if it was the antenna that caused the change in reception. so he wasn't only measuring the reception in his house he was measuring it based on different antennas
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The land between two normal faults moves upward to form a
Leya [2.2K]
<span>The land between two normal faults moves upward to form a

Answer:D</span><span>
fault-block mountain.</span>
5 0
3 years ago
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If it takes Ashley 6.1 seconds to run at an average speed of 5.7 meters per second, what is the distance (in meters) she covers
Softa [21]

Answer:

34.8m

Explanation:

distance = speed x time

6.1 x 5.7 = 34.77

34.8

3 0
3 years ago
A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

5 0
3 years ago
Read 2 more answers
A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

Mass of the climber = 69.38 kg

6 0
3 years ago
An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object
tatiyna

Answer:

D_T=18.567m

Explanation:

From the question we are told that:

Acceleration a=8.0 m/s^2

Displacement d=1.05 m

Initial time t_1=6.0s

Final Time t_2=2.5s

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

 V^2=2as

 V=\sqrt{2*6*1.05}

 V=4.1m/s

Generally the equation for Distance traveled before stop is mathematically given by

 d_2=v*t_1

 d_2=3.098*4

 d_2=12.392

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity v_3=0 m/s

Initial velocity u_3=4.1 m/s

Therefore

Using Newton's Law of Motion

 -a_3=(4.1)/(2.5)

 -a_3=1.64m/s^2

Giving

 v_3^2=u^2-2ad_3

Therefore

 d_3=\frac{u_3^2}{2ad_3}

 d_3=\frac{4.1^2}{2*1.64}

 d_3=5.125m

Generally the Total Distance Traveled is mathematically given by

 D_T=d_1+d_2+d_3

 D_T=5.125m+12.392+1.05 m

 D_T=18.567m

6 0
2 years ago
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