Do Earth’s oceans gain or lose water, considering evaporation and precipitation together? How much?

. (a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy bar? (b) Does this seem like a long time?

galina1969 [7]

(a). The power of the candy bar is,

$P=440\text{ W}$

The time taken to play on 800 kJ energy of the candy bar is,

$t=\frac{E}{P}$

where E is the energy.

Substituting the known values,

$\begin{gathered} t=\frac{800\times10^3}{440} \\ t=1.818\times10^3\text{ s} \\ t=1818\text{ seconds} \end{gathered}$

As 1 minute is equal to 60 seconds,

Thus,

$\begin{gathered} t=1818\text{ seconds} \\ t=\frac{1818}{60} \\ t=30.3\text{ min} \end{gathered}$

Thus, the time taken to play tennis on the 800 kJ energy is 30.3 minutes.

(b). By doing the exercise, the process of digestion of food inside our body increases. Thus, the exercise does not helps us to burn the calories. But it helps us to diggest the heavy meal like candy bar easily.

The time taken to digest the canndy bar or to utilise its energy is large because it takes a lot of time to burn small amount of food and make it digest quickly.

8 0

1 year ago

What is the change in momentum of a 1000kg car moving from 1 m/s to 8m/s

natita [175]

Answer:

Explanation:

In order to calculate the change of any quantity you must substract the initial value of that quantity from the final value of the same quantity.

The concept of momentum is defined as,

Momentum = mass×velocity

So we calculate the change of momentum as,

Momentum change = final momentum - initial momentum

= mv - mu =m(v-u)

Where v = final velocity

u = initial velocity

m = mass of the object

So to calculate the momentum change of the car,

Change in momentum of the car = 1000×(8-1) = 1000×7 = 7000 Ns

5 0

3 years ago

A wagon with an initial velocity of 2 m/s and a mass of 60 kg, gets a push with 150 joules of

Aleks04 [339]

Answer:

v_f = 3 m/s

Explanation:

From work energy theorem;

W = K_f - K_i

Where;

K_f is final kinetic energy

K_i is initial kinetic energy

W is work done

K_f = ½mv_f²

K_i = ½mv_i²

Where v_f and v_i are final and initial velocities respectively

Thus;

W = ½mv_f² - ½mv_i²

We are given;

W = 150 J

m = 60 kg

v_i = 2 m/s

Thus;

150 = ½×60(v_f² - 2²)

150 = 30(v_f² - 4)

(v_f² - 4) = 150/30

(v_f² - 4) = 5

v_f² = 5 + 4

v_f² = 9

v_f = √9

v_f = 3 m/s

7 0

3 years ago

A web page designer creates an animation in which a dot on a computer screen has a position of r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i

VMariaS [17]

Answer:

V = (5.8cm/s)i, (4.7cm/s)j

Explanation:

Given :

r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^

To obtain the average velocity (V)

V = (r2 - r1) / (t2 - t1)

To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above

r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j

r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j

r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j

r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j

r2 = (16.1cm)i + (9.4cm)j

V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0

V = 11.6i / 2 ; 9.4j / 2

V = (5.8cm/s)i, (4.7cm/s)j

5 0

3 years ago

An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet

Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

Object 3 should be placed closer to object 1.

Object 3 should be placed on a halfway between object 2 and object 1.

Object 3 should be placed closer to object 2.

Solution

I think that Object 3 should be placed closer to object 2.

6 0

3 years ago