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Vesnalui [34]
2 years ago
9

Suppose you are in an elevator that is moving upward with a constant velocity. A scale inside the elevator shows your weight to

be 600 N.
(a) Does the scale register a value that is greater than, less than, or equal to 600 N during the time when the elevator slows down as it comes to a stop?
(b) What is the reading when the elevator is stopped?
(c) How does the value registered on the scale compare to 600 N during the time when the elevator picks up speed again on its way back down? Give your reasoning
Physics
1 answer:
Ulleksa [173]2 years ago
4 0

Answer:

a) Less than 600 N

b) 600 N

c) Less than 600 N

Explanation:

The value registered by the scale, measures the normal force on the object placed on the scale.

The person on the scale, is acted by two external forces, gravity (which we call weight, and always go downward) and the normal force, an upward force in this case.

The difference between both forces, according Newton’s 2nd law, must be equal to the product of the mass of the object, times the acceleration.

If the elevator moves upward to a constant speed, and then slows down, this means that there exists a net force on the person, producing a acceleration directed downwards on him.

If we take the direction of the acceleration (downward) to be positive, this means that the difference between gravity force and normal force must be positive.

As mg = 600 N (equal to normal force when no net force is present), the normal force in this case must be less than 600 N.

b) If the elevator is stopped, the effect is the same like it were moving at constant speed, so in this case the normal force remains the same: 600 N

c) Assuming that it starts from rest, if it accelerates going downward, and if we take as positive the downward direction, in order to satisfy Newton’s 2nd Law, normal force must be less than 600N so the difference between gravity and normal force remain positive.

You might be interested in
Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme
Artyom0805 [142]

Heat engines were developed during industrial revolution.

Generally a heat engine contains three parts i.e source, sink and working substance.

The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider  T_{1}\ and\ T_{2} are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

Let\ Q_{1}\ and\ Q_{2} are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

                                                W\ =\ Q_{1}-Q_{2}

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

                     Efficiency\ [\eta]=\frac{W}{Q_{1}}

                                        \eta\ =\frac{Q_{1}- Q_{2}} {Q_{1}}

                                               =\ 1-\frac{Q_{2}} {Q_{1}}

This is the expression for the efficiency of heat engine.

Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.

6 0
2 years ago
Read 2 more answers
A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla
ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

4 0
2 years ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
2 years ago
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
lions [1.4K]

Answer:

a) Probability mass function of x

x P(X=x)

0 0.0602

1 0.0908

2 0.1700

3 0.2050

4 0.1800

5 0.1550

6 0.0843

7 0.0390

8 0.0147

b) Cumulative Distribution function of X

x F(x)

0 0.0602

1 0.1510

2 0.3210

3 0.5260

4 0.7060

5 0.8610

6 0.9453

7 0.9843

8 1.0000

The cumulative distribution function gives 1.0000 as it should.

Explanation:

Probability of arriving late = 0.43

Probability of coming late = 0.57

Let's start with the probability P(X=0) that exactly 0 people arrive late, the probability P(X=1) that exactly 1 person arrives late, the probability P(X=2) that exactly 2 people arrive late, and so on up to the probability P(X=8) that 8 people arrive late.

Interpretation(s) of P(X=0)

The two singles must arrive on time, and the three couples also must. It follows that P(X=0) = (0.57)⁵ = 0.0602

Interpretation(s) of P(X=1)

Exactly 1 person, a single, must arrive late, and all the rest must arrive on time. The late single can be chosen in 2 ways. The probabiliy that (s)he arrives late is 0.43.

The probability that the other single and the three couples arrive on time is (0.57)⁴

It follows that

P(X=1) = (2)(0.43)(0.57)⁴ = 0.0908

Interpretation(s) of P(X=2)

Two late can happen in two different ways. Either (i) the two singles are late, and the couples are on time or (ii) the singles are on time but one couple is late.

(i) The probability that the two singles are late, but the couples are not is (0.43)²(0.57)³

(ii) The probability that the two singles are on time is (0.57)²

Given that the singles are on time, the late couple can be chosen in 3 ways. The probability that it is late is 0.43 and the probability the other two couples are on time is (0.57)².

So the probability of (ii) is (0.57)²(3)(0.43)(0.57)² which looks better as (3)(0.43)(0.57)⁴ It follows that

P(X=2) = (0.43)²(0.57)³ + (3)(0.43)(0.57)⁴ = 0.0342 + 0.136 = 0.1700

Interpretations of P(X=3).

Here a single must arrive late, and also a couple. The late single can be chosen in 2 ways. The probability the person is late but the other single is not is (0.43)(0.57).

The late couple can be chosen in 3 ways. The probability one couple is late and the other two couples are not is (0.43)(0.57)². Putting things together, we find that

P(X=3) = (2)(3)(0.43)²(0.57)³ = 0.2050

Interpretation(s) P(X=4)

Since we either (i) have the two singles and one couple late, or (ii) two couples late. So the calculation will break up into two cases.

(i) Two singles and one couple late

Two singles' probability of being late = (0.43)² and One couple being late can be done in 3 ways, so its probability = 3(0.43)(0.57)²

(ii) Two couples late, one couple and two singles early

This can be done in only 3 ways, and its probability is 2(0.57)³(0.43)²

P(X=4) = (3)(0.43)³(0.57)² + (3)(0.57)³(0.43)² = 0.0775 + 0.103 = 0.1800

Interpretations of P(X=5)

For 5 people to be late, it has to be two couples and 1 single person.

For couples, The two late couples can be picked in 3 ways. Probability is 3(0.43)²(0.57)

The late single person can be picked in two ways too, 2(0.43)(0.57)

P(X=5) = 2(3)(0.43)³(0.57)² = 0.1550

Interpretations of P(X=6)

For 6 people to be late, we have either (i) the three couples are late or (ii) two couples and the two singles.

(i) Three couples late with two singles on time = (0.43)³(0.57)²

(ii) Two couples and two singles late

Two couples can be selected in 3 ways, so probability = 3(0.43)²(0.57)(0.43)²

P(X=6) = (0.43)³(0.57)² + 3(0.43)⁴(0.57) = 0.0258 + 0.0585 = 0.0843

Interpretation(s) of P(X=7)

For 7 people to be late, it has to be all three couples and only one single (which can be picked in two ways)

P(X=7) = 2(0.57)(0.43)⁴ = 0.0390

Interpretations of P(X=8)

Everybody had to be late

P(X=8) = (0.43)⁵ = 0.0147

6 0
2 years ago
An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2

Now we can apply Newton's second law to find the net force on the object:

F=ma=(24 kg)(4 m/s^2)=96 N

This net force is the resultant of the applied force forward (F_a) and the component of the weight acting backward (mg sin \theta), so we can find what is the applied force:

F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N

where

m = 24 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

W=F_a s = (237.6)(18)=4276 J

where s = 18 m is the displacement.

Therefore the average power needed is:

P=\frac{W}{t}=\frac{4276}{3}=1425 W

b)

The instantaneous power at any point of the motion is given by

P=F_av

where

F_a is the force applied

v is the velocity of the object

We already calculated the applied force:

F_a=237.5 N

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

v=u+at=0+(4)(3.0)=12.0 m/s

And therefore, the instantaeous power at 3.0 sec is:

P=Fv=(237.5)(12)=2850 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
2 years ago
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