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3 years ago

Why does it takes a long time to heat a room with a high ceiling?

Physics

2 answers:

vlada-n [284]3 years ago

6 0

Answer:

High ceilings make a room feel large and open, but they can be difficult to cool and heat. Because hot air rises, the challenge becomes trying to keep the hot air where you want it and preventing if from being wasted where you don't.

Explanation:

ddd [48]3 years ago

3 0

Answer:High ceilings make a room feel large and open, but they can be difficult to cool and heat. Because hot air rises, the challenge becomes trying to keep the hot air where you want it and preventing if from being wasted where you don't

Explanation:

You might be interested in

Suppose a car travels at a constant speed of 12.0 m/s. How far would it move in 3600.0 s?

beks73 [17]

Answer:

43200 m

Explanation:

speed = 12.0 m/s

time = 3600.0 s

distance = speed * time

distance = 12.0 m/s * 3600.0 s

distance = 43200 m

7 0

3 years ago

If the merry-go-round makes one revolution in 10 seconds, what is the child’s linear speed?

Anton [14]

The child's linear speed is

<em> (pi / 5) x (the child's distance from the center of the ride, in feet)</em>

feet per second.

5 0

3 years ago

A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the vel

spin [16.1K]

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

$m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s$

So Option d is correct one

5 0

3 years ago

Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia

Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km west

velocity V1 = 8.9 km/h

distance B = 4.5 km east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7 ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x ) / 8.9 .............2

now equating equation 1 and 2

time A = time B

x / 7 = ( 10.2 - x ) / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0

3 years ago

I. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maxi

nalin [4]

Answer:

Negative

Explanation:

If the box is heading right in the positive direction, the work will be negative. The spring has an opposite force to that of the box.

Hope this helped. :)

6 0

3 years ago