HELPPPP

¿De qué altura cae un cuerpo que tarda 4 s en llegar al suelo? usar g = 10 m/s2 *

jasenka [17]

Answer:

hi

Explanation:

3 0

3 years ago

A 7.0kg skydiver is descending with a constant velocity

Vikentia [17]

Answer:

The air resistance on the skydiver is 68.6 N

Explanation:

When the skydiver is falling down, there are two forces acting on him:

- The force of gravity, of magnitude $mg$ , in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)

- The air resistance, $R$ , in the upward direction

So the net force on the skydiver is:

$F=mg-R$

where

m = 7.0 kg is the mass

$g=9.8 m/s^2$

According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):

$F=ma$

In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:

$a=0$

Therefore the net force is zero:

$F=0$

And so, we have:

$mg-R=0$

And so we can find the magnitude of the air resistance, which is equal to the force of gravity:

$R=mg=(7.0)(9.8)=68.6 N$

6 0

4 years ago

In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres o

SVETLANKA909090 [29]

Answer:

$F = 7.7*10^{-10}N$

Explanation:

You need to be careful with units for this problem. The force will be:

$F =\frac{K*m1*m2}{d^2}$

$F=\frac{6.67259 * 10^{-11}*1.56*21.1*10^{-3}}{(5.34*10^{-2})^2}$

$F=7.7*10^{-10}N$

8 0

3 years ago

Which type of mirror has a flat surface?

Vsevolod [243]

Most mirrors are plane mirrors that have a flat reflective surface. A plane mirror forms only virtual, right-side up, and life-sized images. A concave mirror is shaped like the inside of a bowl.

3 0

3 years ago

Read 2 more answers

I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!

sergij07 [2.7K]

Answer:

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

V = 12 V
P = 24 W

$\implies \mathsf{24=\frac{12^2}{R} }$

$\implies \mathsf{24R=12^2 }$

$\implies \mathsf{24R=144 }$

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

V = 240 V

$\implies \mathsf{100=\frac{240^2}{R} }$

<em>R and 100 can interchange places</em>

$\implies \mathsf{R=\frac{240^2}{100} }$

$\implies \mathsf{R=\frac{57600}{100} }$

<u></u>

By Ohm's Law:

$\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}$

=> 240 = I × 576

=>

=> I = 0.417 A

I don't know it's resistance,... so sorry

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

$\boxed{\mathfrak{Power \propto \frac{1}{Resistance} }}$

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

3 0

3 years ago