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tresset_1 [31]
3 years ago
11

Anna is conducting an experiment to determine how weather affects cell phone reception. She is trying to decide the best way to

conduct her experiment in order to collect meaningful data. Which of the following experiments would help Anna collect the best data? A. Test different cell phones in different locations on days with clear weather. B. Test different cell phones in different locations on days with rainy weather. C. Test a cell phone's reception in one location with clear weather and in another location with rainy weather. D. Test a cell phone's reception in the exact same location under various atmospheric conditions.
Physics
1 answer:
aev [14]3 years ago
4 0

Answer:

D. Test a cell phone's reception in the exact same location under various atmospheric conditions

Explanation:

Components of an experiment:

Independent variable:

A manipulated variable, in an experiment or study, whose presence or degree incurs a change in the  dependent variable.

Dependent variable:

A variable which is being studied in the experiment and is supposed to change with respect to the independent variable.

Control variable:

Control variable includes the variables which are of no concern in the experiment and must be constant so that they do not interfere with the results of the experiment. Change in control variable might affect the results of an experiment.

In Anna's experiment, the independent variable is the atmospheric conditions. The dependent variable is cell phone reception that will change according to the atmospheric conditions. All other things related to the experiment are control variables like cellphone, location of the cell phone. Both of these must stay constant to collect the best data.

Hence, the best option is D.

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A 80 ohms resistor, 0.2 H inductance and 0.1 mF capacitor are connected in series across a generator (60 Hz, V rms=120 V). Deter
qaws [65]

Answer:

Impedance = 93.75 ohms

Current = 1.81 A

Explanation:

Resistance = R = 80 ohms

Inductance = L = 0.2 H

Inductive reactance = XL = X_{L}= = ωL = (2πf) L

= 2 (3.14) (60)(0.2) = 75.398 Ohms

Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]

= 26.526 Ohms  

Impedance = Z = \sqrt{R^{2} + (X_{L} - X_{C})^{^{2}}} =

= \sqrt{8788.511} = 93.747 ohms  

Voltage = \sqrt{2} × 120 = 169.7056 V

Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A

8 0
3 years ago
A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi
Marianna [84]

Answer:

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

8 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr
mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v²   ........1

6.5 × 1.6 × 10^{-19} = 1/2 × 1.672 × 10^{-27} ×v²

v = 3.5 × 10^{7} m/s

so

radius will be

radius = \frac{m*v}{B*q}   ........2

radius =  \frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}  

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

8 0
3 years ago
Read 2 more answers
Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.
mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, \rho=10^{-6}\ \Omega -m

Density, d=8.31\times 10^3\ kg/m^3

(A) The length of the wire is given by using the definition of resistance as :

Volume,

V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3

Area,

A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}....(1)

R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m

(b)  Equation (1) becomes :

A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m

Hence, this is the required solution.                                                                  

5 0
3 years ago
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