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3 years ago

The potential energy of a 60 kg man standing on a balcony 43 meters above the ground is:

2 answers:

3 0

Given: Mass m = 60 Kg; Height h = 43 m

Required: Potential P.E

Formula: P.E = mgh

P.E = (60 Kg)(9.8 m/s²)(43 m)

P.E = 25,284 J

ArbitrLikvidat [17]3 years ago

3 0

Connexus Physics pre-ap semester A final exam, all correct answers. And y’all better be freaking grateful, because it took a long time to type all this out.

1) In the steps of the scientific method…
A. Interpreting results

2) The radius of the Earth is…
B. 6.37 x 10^3 km

3) Which of the following equations best describes…
D. y = 1/2x

4) In the graph, what is the correct description…
C. Negative position

5) The motion of a ball on an inclined plane is described…
C. vi

6) Acceleration due to gravity is also called
C. free-fall acceleration

7) Which of the following is a physical quantity…
A. vector

8) In a coordinate system, a vector is…
A. Cos (theta)

9) Which of the following is an example of projectile motion?
B. a thrown baseball

10) A newton is equivalent to which…
A. kg * m/s^2

11) In the free-body diagram shown below, which of the following…
C. 4050 N

12) A net force of 6.8 N accelerates a 31 kg scooter…
A. 0.22 m/s^2

13) A change in the gravitational force acting…
C. Weight

14) A passenger on a bus moving east sees a man standing on a curb…
C. move west at a speed that is equal to the bus’s speed.

15) In which of the following scenarios is no net work done?
B. A car travels at a constant speed on a flat road.

16) What is the kinetic energy of a 0.135 kg baseball thrown at 40.0 m/s?
C. 108 J

17) Which of the following refers to the sum of kinetic energy…
D. mechanical energy

18) Which of the following are not units of power?
B. J

19) Which of the following…can be used to…calculate…change in…momentum?
D. (delta)p = F (delta) t

20) Two skaters stand facing…
B. their momenta are equal but opposite.

21) In an inelastic collision…
D. the momentum of one object will increase by the amount that the momentum of the other object decreases.

22) When an object is moving with uniform circular motion…tangential speed
C. is constant

23) Which units would be the most appropriate to use to…
A. newtons

24) How would the speed of Earth’s orbit around the sun change…
A. The speed would not change

25) A girl pushes a box that has a weight of 450 N up an incline…force of 150 N along…
B. 3.0

26) According to the law of conservation of mechanical energy,
C. Total energy must remain the same

27) The potential energy of a 60 kg man…balcony 43 meters…
A. 25000 J

28) If two small beakers of water, one at 70C and one at 80C…
C. The final temperature is between 70C and 80C

29) Why does sandpaper get hot when it is rubbed against rusty metal?
C. Friction between the sandpaper and metal…

30) Which of the following describes a substance in which the temperature and pressure…
D. substance undergoing a change of state

31) An ideal gas system is maintained at a constant volume of 4 L. If pressure is constant…
A. 0 J

32) Which of the following is a thermodynamic process in which a system…
C. a cyclic process

33) When a drop of ink mixes with water…
A The system’s entropy increases, and the total entropy of the universe increases.

34) A mass-spring system can oscillate…
D. elastic potential

35) If a semi-truck decrease its speed from 10 m/s to 5 m/s...
D. ¼ its old kinetic energy

36) A boy releases a small 1 kg toy truck from rest down a 5 meter high hill…
B. 50 J

37) A car accelerates uniformly from 10 m/s to 20 m/s in 4.2 seconds…
A. 2.4 m/s^2

38) What must the minimum tension of a single vertical wire be to support a 1.5 kg picture?
A. 15 N

39) Why does it seem harder to start sliding a couch across a carpeted floor than…
C. The opposing force of friction is greater when an object starts from rest…

40) An elevator weighing 2.00 c 10^5 N is supported by a steel cable…
Fw = mass * gravitational acceleration
2*10^5 = mass * 9.8
mass of the elevator = 20.4*10^3 kg

Fnet = Tension – Fw
mass * acceleration = Tension – 2*10^5
20.4*10^3 * 3 = Tension – 2*10^5
61200 = Tension – 2*10^5
Tension = (61200) + (2*10^5)
Tension = 2.61 * 10^5 newton

41) A bullet with a mass of 6.00 g is fired through a 1.25 kg block of wood on a frictionless surface.
0.006(896) + 0 = 0.006(435) + 1.25(V)
5.376 + 0 = 2.61 + 1.25(V)
5.376 = 2.61 + 1.25(V)
2.766 = 1.25(V)
V = 2.21 m/s

42) Find the final equilibrium temperature when 10.0 g of milk at 10.0C is added to 1.60 x 10^2 g of coffee with a temperature of 90.0C.
Heat lost = heat gained
X = equilibrium temperature
Heat lost = 160(90-x)
Heat gained = 10(x-10)
14400 – 160x = 10x -100
170x = 14500
X = 14500/170
X = 85.25 degrees

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

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Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

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3 years ago

In moving out of a dormitory at the end of the semester, a student does 1.82 x 104 J of work. In the process, his internal energ

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Answer:

Explanation:

(a) Work done, W = 1.82 x 10^4 J

(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)

Q = W + U

Q = 1.82 x 10^4 - 4.07 x 10^4

Q = - 2.25 x 10^4 J

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3 years ago

Science help answer all

wlad13 [49]

Idk but i hope you figure it out :)

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3 years ago

What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an

astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution= $T_b=101^{\circ}$ C

Boiling point water=100 degree Celsius

$K_f=1.86K/m$

$K_b=0.512 K/m$

$\Delta T_b=T-T_0$

Where $T$ =Boiling point of solution

$T_0=$ Boiling point of pure solvent

$\Delta T_b=101-100=1^{\circ}$ C

$\Delta T_b=k_bm$

Using the formula

$1=0.512\times m$

Molality, $m=\frac{1}{0.512}$ m

$\Delta T_f=k_fm$

Using the formula

$\Delta T_f=\frac{1}{0.512}\times 1.86$

$\Delta T_f=3.63 C$

We know that

$\Delta T_f=T_0-T_1$

Where $T_0$ =Freezing point of solvent

$T_1=$ Freezing point of solution

Using the formula

$3.63=0-T_1$

Freezing point of water=0 degree Celsius

$T_1=0-3.63=-3.63 C$

Hence, the freezing point of solution=-3.63 degree Celsius

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Explanation:

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3 years ago