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andre [41]
4 years ago
5

A student must determine the relationship between the inertial mass of an object, the net force exerted on the object, and the o

bject’s acceleration. The student uses the following procedure. The object is known to have an inertial mass of 1.0kg .
Step 1: Place the object on a horizontal surface such that frictional forces can be considered to be negligible.

Step 2: Attach a force probe to the object.

Step 3: Hang a motion detector above the object so that the front of the motion detector is pointed toward the object and is perpendicular to the direction that the object can travel along the surface.

Step 4: Use the force probe to pull the object across the horizontal surface with a constant force as the force probe measures force exerted on the object. At the same time, use the motion detector to record the velocity of the object as a function of time.

Step 5: Repeat the experiment so that the object is pulled with a different constant force.

Can the student determine the relationship using this experimental procedure?

Answer choices:

A) Yes, because Newton’s second law of motion must be used to determine the acceleration of the object.

B) Yes, because the net force exerted on the object and its change in velocity per unit of time are measured.

C) No, because the motion detector should be oriented so that the object moves parallel to the line along which the front of the motion detector is aimed.

D) No, because knowing the net force exerted on the object and its change in velocity per unit of time is not sufficient to determine the relationship.
Physics
1 answer:
Effectus [21]4 years ago
3 0

Answer:

C

Explanation:

In order to obtain data about the object’s velocity as a function of time, the object must move either toward or away from the motion detector.

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A simple generator has a square armature 5.0 cm on a side. The armature has 95 turns of 0.59-mm-diameter copper wire and rotates
guajiro [1.7K]

Answer:

25.39Hz

Explanation:

#First, we need to determine the actual emf required. The generator's internal resistance will cause a voltage drop inside the generator/

Internal resistance is defined using the formula:

R=\rho L/A\\\rho=1.68\times10^-^8 \Omega m\\L=0.050m\times4\times 60=12.0m\\A=\pi r^2=\pi d^2/4=\pi(0.00059m)^2/4=2.734\times10^-^7m^2\\\\R=(1.68\times10^-^8 \Omega \ m\times 12.0m)/2.734\times10^-^7m^2\\=0.7374\ \Omega

#The bulb is rated 12.0V,25.0W

Current, I=25.0W/12.0V=2.083A

Therefore, the voltage drop in the generator is calculated as:

2.083A\times0.7374\Omega=1.5360V  

Actual EMF required is thus   1.536V+12.0V=13.536V

#peak voltage is 13.536V\sqrt 2=19.143V

#For a generator, by Faraday's Law

E_m_a_x=NBA\ \omega\\19.143=60\times 0.800T\times (0.05m)^2\ \omega\\\\\omega=159.525rad/s

f=\omega/2\pi\\=159.525/2\pi=25.39Hz

#The rate of the generator is 25.39Hz

7 0
4 years ago
Convert 5 kilograms to pounds show your work
lyudmila [28]

Answer:

11

Explanation:

There are 2.2lbs in a kilo

times the KG by 2.2 and u get the amount in lbs (pounds)

6 0
3 years ago
The alternating current which crosses an apparatus of 600 W has a maximum value of 2.5 A. What is efficient voltage between its
azamat

Answer: Option (b) is correct.

Explanation:

Since we know that,

P = VI

where;

P = power

V= Voltage

I = Current

Since it's given that,

P = 600W

I = 2.5 A

equating these values in the above equation, we get;

<em>V = \frac{600}{2.5}</em>

<em>V = 240 V</em>

8 0
3 years ago
What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicl
Lapatulllka [165]

Answer:

276.74\times 10^8Mg/m^3

31.29 m/sec

Explanation:

We have given density of substance 0.14lb/in^3

We have convert this into Mg/m^3

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram =63.49\times 10^3\ Mg

We know that 1 in^3=1.6387\times 10^{-5}m^3

So 0.14in^3=0.14\times 1.6387\times 10^{-5}=0.2294\times 10^{-5}m^3

So 0.14lb/in^3 =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr =\frac{70\times 1609.34meter}{3600sec}=31.29m/sec

8 0
3 years ago
Scenario:The owner of Bond's Gym wants your advice. He asks you if you think positive incentives would work better than negative
kakasveta [241]

Answer:

1. Negative Incentives

2. Decrease

3. More

Explanation:

Edge 2021

3 0
3 years ago
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