Ask question

Ask question

All categories

3 years ago

8

17. What is the gain in gravitational potential energy of a body of weight 2000 N as it rises from a height of 20 m to a height

of 25 m above the earth's surface?
(A) 400 J
(B) 1 000 J
(C) 10 000 J
(D) 20 000 J

1 answer:

Tju [1.3M]3 years ago

7 0

Answer:

C) 10000 J

Explanation:

∆p.e =mg∆h

= 2000 × 5

= 10000J

You might be interested in

Which of the following statements best describes earths mineral resources

bulgar [2K]

It is D, you can not replace minerals. This makes them valuable.

8 0

3 years ago

Read 2 more answers

A person travels by car from one city to another with different constant speeds between pair of cities. She drives for 36 min at

Softa [21]

Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
:

let a = average speed for the trip
:
Write a dist equation, dist = speed * time
:
80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance

3 0

3 years ago

When jogging outside you accidently bump into a curb. Your feet stop but your body continues to move forward and you end up on t

oee [108]

Inertia I think because I've heard it around school and in science

6 0

3 years ago

A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?

ipn [44]

Your answer would be A. You divide 96 by 16 to find the answer

3 0

4 years ago

Read 2 more answers

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

3 years ago