PHYSICS COLLEGE

Let A be the last two digits, and let B be the last three digits, and the C be the sum of the last 4 digits of your 8-digit student ID. (Example: For 20245347, A = 47, B = 347, and C = 19) A train moves at an average speed of (23.0 + A) m/s for (250.0 + B) seconds and then at an average speed of (45.0 + C) m/s for (800.0 + B) seconds. Determine the average speed for the entire time in meters per second (m/s). Round your final answer to 3 significant figures.

Answers

Answer 1

Answer:

66.053m/s

Explanation:

A = 47

B = 347

C = 19

Train moves at

(23 + A)m/s

= 23 + 47 = 60m/s

At (250.0+B) seconds

250.0+347 =

547 seconds

Distance d,

= 70 x 597

= 41790

It also moves at

(45.0 + c)

= 45 + 19

= 64m/s

Time = 800 + B

= 800 + 347

= 1147

Distance,

= 64 x 1147

= 73408m

Total distance,

= 73408 + 41790

= 115,198

Total time,

= 597 + 1147

= 1744

Average speed,

= Total distance / total time

= 115198/1174

= 66.053m/s

Answer 2

Answer:

Final answer:

The average speed over the entire time can be calculated by first finding the distances the train travels over both periods, then finding the total distance and the total time, and finally dividing the total distance by the total time. The value must be rounded to three significant figures.

Explanation:

You can find the average speed of the train over the full-time interval by dividing the total distance travelled by the total time. To begin with, you would have to find the distances the train covered during both periods.

The distance (D1) it travelled during the first period can be found by multiplying the average speed (23.0 + A) by the time (250.0 + B).
The distance (D2) it travelled during the second period can be calculated by multiplying the average speed (45.0 + C) by the time (800.0 + B).

Then you add D1 and D2 to get the total distance (TD). This will be (D1 + D2). The total time (TT) will be found by adding both time intervals, which means it equals (250.0 + B) + (800.0 + B). You then divide the total distance by the total time to get the average speed, i.e., TD/TT. Lastly, round the average speed to 3 significant figures.

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What is the magnitude of a vector that has the following components: x = 32 m y = -59 m

Answers

Answer:

Explanation:

Since the x and y components are given

The vectors Magnitude = √32²+(-59)²

=67.12m

7. An engineer is using a wire that has a resistance of 1.5 . This resistance is too high for the application he is designing. The wire must be exactly 2.5 cm long. What two things could he do to reduce the wire's resistance

Answers

Answer and Explanation:

We know that resistance $R=(\rho l)/(A)$ from the given equation of resistance it is clear that resistance depends on resistivity length and area of the material but we can not change the length because it is given that the length must be 2.5 cm long.

So we can do two two things to reduce the resistance

increase the cross sectional area
decrease the resistivity of the material

The mass of a string is 20 g and it has a length of 3.2 m. Assuming that the tension in the string is 2.5 N, what will be the wavelength of a travelling wave that is created by a sinusoidal excitation of this string with a frequency of 20 Hz. Provide the wavelength in units of m. Please note: You do not include the units in your answer. Just write in the number.

Answers

Answer:

The wavelength of the wave is 1 m

Explanation:

Given;

mass of the string, m = 20 g = 0.02 kg

length of the string, L = 3.2 m

tension on the string, T = 2.5 N

the frequency of the wave, f = 20 Hz

The velocity of the wave is given by;

$v = \sqrt(T)/(\mu) {}$

where;

μ is mass per unit length = 0.02 kg / 3.2 m

μ = 6.25 x 10⁻³ kg/m

$v = \sqrt{(T)/(\mu) } \n\nv = \sqrt{(2.5)/(6.25*10^(-3)) } \n\nv = 20 \ m/s$

The wavelength of the wave is given by;

λ = v / f

λ = (20 m/s )/ (20 Hz)

λ = 1 m

Therefore, the wavelength of the wave is 1 m

An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a less dense liquid. What would you observe?

Answers

Answer:

The fraction of its volume inside liquid is increased .

Explanation:

According to principle pf floatation , an object floats on the surface of water

when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .

In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .

A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°C. What will be the volume of the air in the bag if you take it with you, still sealed, to the mountains where the temperature is 7.00°C and atmospheric pressure is 70.0 kPa

Answers

The volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

What is the gas law?

The gas law is used to show the relationship between the pressure and the temperature of the gases. It can be given as,

$PV=nrT$

Here, (n) and (r) are the constant. Therefore,

$(PV)/(T)=\rm Constant$

For the initial and final values, the gas law can be given as,

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$

Here, (subscript 1,and 2) is used for the initial and final amount of pressure and temperature.

The initial values of the bag of potato chips as volume of 2.00 L, pressure of 1.00 ATM and a temperature of 20.0°C. It is known that the value of 1 ATM is equal to the 101.325 kPa.

The final temperature of the pack is 7.00°C and atmospheric pressure is 70.0 kPa

Put the values in the above formula as,

$(101.325*2)/(293)=(70* V_2)/(280)\nV_2=2.766\rm \; liters$

Hence, the volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

Learn more about the gas law here;

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Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_(1) = 2.00\ L$

Pressure $P_(1)= 1.00\ atm$

Pressure $P_(2)= 70.0\ kPa$

Temperature $T_(1)= 20.0°C = 293\ K$

Temperature $T_(2)= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$(PV)/(T)=\ Constant$

For both temperature,

$(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))$

Put the value into the formula

$(101.325*2)/(293)=(70* V_(2))/(280)$

$V_(2)=(101.325*2*280)/(293*70)$

$V_(2)=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

According to American Heart Association, your target heart rate is__________. Group of answer choices

A .all of the above

B. 85% - 95% of your max heart rate

C. 20%-30% of your max heart rate

D. 50%-85% of your max heart rate

Answers

Answer:

Explanation:

A normal heart rate is from 85-95% the other heart rate is not normal because your heart is beating more than normal

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