Why can't a hot air balloon go to higher layers?

I need help with questions b and d, that’s all. Thank you.

Mkey [24]

b). The power depends on the RATE at which work is done.

Power = (Work or Energy) / (time)

So to calculate it, you have to know how much work is done AND how much time that takes.

In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.

The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.

Without knowing the time, we can't calculate the power.

...

d). Kinetic energy = (1/2) · (mass) · (speed squared)

On the way up, the car stops when it reaches point-A.

On the way down, the car leaves point-A from "rest".

WHILE it's at point-A, it has no speed. So it has no (zero) kinetic energy.

7 0

3 years ago

A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano $(m)$ = 190 kg

Inclined angle $(\theta)$ = 18 degree

Considering gravity, $g$ = 9.8 $ms^-^2$

And

Using, $sin(18) =0.30$ and $cos(18)=0.95$

FBD diagram is attached with all the force acting on the floor and and the inclined.

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as $\sum F=0$ .

⇒ $F+mgsin(\theta) =0$

⇒ $F=-mgsin(\theta)$

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass $(m)$ and angle $(\theta)$ .

⇒ $F=190\times 9.8\times sin(18)$

⇒ $F=575.38$ N

b.

When the force is parallel to the floor.

⇒ $Fcos(\theta)=mgsin(\theta)$

⇒ $F=\frac{mgsin(\theta)}{cos(\theta)}$

⇒ Plugging the values.

⇒ $F=\frac{190\times 9.8\times sin(18)}{cos(18)}$

⇒ $F=605$ N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

6 0

3 years ago

How much work is done on a pumpkin with a force of 16 newtons when you lift it 15 meters?

blsea [12.9K]

Answer:

240 Joules

Explanation:

work = f x d

16 newtons is the force

15 meters is the distance

15*16= 240

3 0

2 years ago

A particular spring has a spring constant of 25 N/m.

kotegsom [21]

Answer:

$e.p.e = \frac{1}{2} k {x}^{2} \\ = \frac{1}{2} \times 25 \times (0.015m)^{2} \\ = 0.0028125$

0.0028125N

8 0

3 years ago

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive rea

Slav-nsk [51]

Answer:

the capacitive reactance is reduced by a factor 2

Explanation:

The capacitive reactance is given by:

$X_C = \frac{1}{2\pi f C}$

where

f is the frequency of the source

C is the capacitance

In this case, the frequency of the source is doubled:

f' = 2f

while the capacitance does not change. So the new capacitive reactance will be

$X_C' = \frac{1}{2\pi f' C}=\frac{1}{2\pi (2f)C}=\frac{1}{2}X_C$

so the capacitive reactance is reduced by a factor 2.

5 0

3 years ago

Why can't a hot air balloon go to higher layers?​

Why can't a hot air balloon go to higher layers?