Ask question

Ask question

All categories

2 years ago

8

Mind being a helping hand? I know ABSOLUTELY nothing about air resistance.. It's still science for me so I guess I'll put it as

physics?

1 answer:

Contact [7]2 years ago

8 0

Answer:

the first one would

Explanation:

air resistance and the second would be gravity or mass

if you need an explanation ask

You might be interested in

A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel

Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0

2 years ago

Does wax paper, aluminum foil, or plastic wrap cause more friction

Levart [38]

Yes i does. they cause more friction <span />

4 0

3 years ago

Read 2 more answers

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil

MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

b)

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!

6 0

3 years ago

What type of lens is used to correct nearsightedness.

Stella [2.4K]

Answer:

Concave lenses are used in eyeglasses that correct myopia or nearsightedness.

5 0

2 years ago

What is the average speed of a cheetah that runs 88 m in 5 seconds

vovangra [49]

So, the average speed of the Cheetah is 17.6 m/s.

<h3>Introduction</h3>

Hello ! I'm Deva from Brainly Indonesia. This time, I will help regarding the average speed. The average speed is obtained from finding the average of the speeds that occur or can be detected from the division between distance and travel time. The average speed can be formulated by :

$\boxed{\sf{\bold{\overline{v} = \frac{s}{t}}}}$

With the following condition :

$\sf{\overline{v}}$ = average speed (m/s)
s = shift or distance objects from initial movement (m)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

s = shift = 88 m
t = interval of the time = 5 seconds

What was asked :

$\sf{\overline{v}}$ = average speed = ... m/s

Step by step :

$\sf{\overline{v} = \frac{s}{t}}$

$\sf{\overline{v} = \frac{88}{5}}$

$\boxed{\sf{\overline{v} = 17.6 \: m/s}}$

So, the average speed of the Cheetah is 17.6 m/s.

5 0

2 years ago