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2 years ago

A heater gives off heat at a rate of 330 kj/min. what is the rate of heat output in kilocalories per hour? (1 cal 4.184 j)

1 answer:

4 0

Answer: 4732.3135 kcal/hour

To answer this question you need to convert the kj into kcalorie then convert the min into hour. Becareful because 1 kilocalories = 1000 calories = 4.184 kjoule.
The calculation would be:
330kj/min x (1kilocalorie/4.184kj) x 60 min/hour= 4732.3135 kcal/hour

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A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a heigh

dlinn [17]

Answer:

a) t = H/v0

b) H = -(v0)²/g

Explanation:

Hi there!

a)The position of the balls can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t.

y0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity.

t = time.

For the ball that is thrown upwards, the initial height is zero, then, the equation can be written as follows:

y = v0 · t + 1/2 · g · t²

The second ball is initially at a height H and the initial velocity is zero. The equation of height for the second ball will be:

y = H + 1/2 · g · t²

When the two balls collide, their height is the same. Then, equalizing both equations we can obtain the time at which they collide:

v0 · t + 1/2 · g · t² = H + 1/2 · g · t²

v0 · t = H

t = H/v0

b) When the first ball is at the highest point its velocity is zero. Using the equation of velocity we can find the time at which the ball is at that point. The equation of velocity is the following:

v = v0 + g · t

At the highest point v = 0.

0 = v0 + g · t

Solving for t:

-v0/g = t

The time at which the first ball is at the highest point is t = -v0/g

The time at which both balls collide was calculated above:

t = H/v0

Then, equalizing both times and solving for H:

H/v0 = -v0/g

H = -v0/g · v0

H = -(v0)²/g

3 0

3 years ago

An electron has ___________

Sauron [17]

Answer:

An electron has a negative electric charge.

3 0

3 years ago

Read 2 more answers

At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops

jasenka [17]

Answer:

force = 11.33 $kg-m/s^{2}$

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

$P_{1}=mU$

$P_{1}= 17.0 * 4.10 = 69.7 kg- m/s$

from newton second law of motion

$F=\frac{\Delta P}{\Delta t}$

$F = \frac{P_{1}-P_{2}}{T}$

Kgm/s^2

$F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}$ [/tex]

8 0

3 years ago

what is the acceleration of an object that moves at a constant velocity of 2.0 meters per second for 5 seconds?

nexus9112 [7]

Constant velocity means moving in a straight line at a speed that doesn't change. If the object is moving with constant velocity then its acceleration is zero. Acceleration is the rate at which velocity is changing.

3 0

3 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

zlopas [31]

Answer:

q = 3.6 10⁵ C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

r = 6 , 37 106 m

E = k q / r²

q = E r² / k

q = $\frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}$

q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

r = 0.6 r_ Earth

r = 0.6 6.37 10⁶ = 3.822 10⁶ m

E = k q / r²

q = E r² / k

q = $\frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}$

q = 3.6 10⁵ C

4 0

3 years ago