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3 years ago

14

When you traveling downhill on a roller coaster, you experience a light feeling in your stomach. why

1 answer:

lesya [120]3 years ago

7 0

Because of the gravitational pull when we go up the gravity pulls u down

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What does flowing electrical charge produce?

Annette [7]

Answer:

Useable energy, electricity

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3 years ago

A bowling ball is on the end of a rope of length 5 meters, the other end of which is attached to a hook in the ceiling. The ball

pogonyaev

Answer:

$a = 0\,\frac{m}{s^{2}}$

Explanation:

Let consider the following system, which is described in the image attached below and two reference axis, one parallel and the other perpendicular to the direction of motion. The corresponding equations of equilibrium are described herein:

$\Sigma F_{n} = T - m\cdot g \cdot \cos \theta = 0$

$\Sigma F_{t} = m\cdot g \cdot \sin \theta = m\cdot a$

The acceleration of the bowling ball at the lowest point occurs at $\theta = 0^{\textdegree}$

$m\cdot g \cdot \sin \theta = m\cdot a$

$g\cdot \sin \theta = a$

$a = 0\,\frac{m}{s^{2}}$

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4 years ago

FIRST TO ANSWER CORRECTLY GETS BRAIBLIEST!!

WINSTONCH [101]

Brushes, battery terminals, commutator, armature, magnets

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3 years ago

Can we store electricity from lightning?

Helga [31]

No. For one thing, when it comes, it flows too fast to catch it and store it. And second, you never know WHERE to set up your equipment.

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3 years ago

Read 2 more answers

How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t

tekilochka [14]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

$Q=mC\Delta T\\\\Q=\rho VC\Delta T$

$(m=\rho V)$

where,

Q = amount of heat required = ?

m = mass

$\rho$ = density of air = $1.20kg/m^3$

V = volume of air

C = specific heat of air = $1006J/kg^oC$

$\Delta T$ = change in temperature = $10.0^oC$

Now put all the given values in above formula, we get:

$Q=\rho VC\Delta T$

$Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)$

$Q=1.449\times 10^6J$

Now we have to calculate the time required.

Formula used :

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

where,

t = time required = ?

Q = amount of heat required = $1.449\times 10^6J$

P = power = 1500 W

Now put all the given values in above formula, we get:

$t=\frac{1.449\times 10^6J}{1500W}$

$t=966s\times \frac{1min}{60s}=16.1min$

Thus, the time required is, 16.1 minutes.

5 0

3 years ago