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PIT_PIT [208]
3 years ago
10

if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball

Physics
1 answer:
grigory [225]3 years ago
3 0
It would exert the same back right?
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A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00
Fantom [35]

Answer:

<h2>The current required  winding is  2.65*10^-^2 mA</h2>

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil  and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= 4\pi*10^-^7 T.m/A

Applying the equation  B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }

I= 2.65*10^-^2 mA

8 0
3 years ago
The drawings show two examples in which a ray of light is refracted at the interface between two liquids. In each example the in
Nataly [62]

Answer:

Possible options:

A. nA, nC, nB

B. nA, nB, nC

C. nC, nB, nA

D. nC, nA, nB

E. nB, nA, nC

Answer is D

Explanation:

When the light is refracted into liquid B it is bent away from the normal, so that nA > nB. When the light is refracted into liquid C it is bent toward the normal, so that nC > nA. Therefore, we conclude that nC > nA > nB

6 0
3 years ago
4 . A negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it . ( a ) what is the unk
Elodia [21]

(a) The unknown charge kept at distance 0.250m directly below  -0.450 charge is  2.315 x 10⁻¹² C.

(b) The magnitude of the force is 29.16 x 10¹² N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

where k = 9 x 10⁹ N.m²/C²

Given is a negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it .

In equation of the magnitude of force, only the magnitude of charges are taken.

Substituting the values into the expression ,we get the charge

0.150 = (9 x 10⁹ x 0.450x q)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the unknown charge is 2.315 x 10⁻¹² C.

b)Let the unknown charge be  0.450C to exert on -450C.

Substituting the values into the expression ,we get the electrostatic force

F = kq₁q₂/r²

F = (9 x 10⁹ x 0.450x 0.450)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the force is 29.16 x 10¹² N.

Learn more about electrostatic force.

brainly.com/question/9774180

#SPJ1

8 0
2 years ago
A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^
Nataliya [291]
Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

E = electric field = 2.86 * 10 ^5 N/C

K = 8.99 * 10^9 N * m^2 / C

d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2  m^2

=> d = 3.07 * 10^-1 m

Answer: 0.307 m

Note: it is a long distance due to the Electric field is very low
7 0
3 years ago
Read 2 more answers
The magnitude of vector λa is 5. Find the value of λ, if: a = (−6, 8)
Nata [24]

Given :

\:  \:

  • \color{pink}{\rm \: The \:magnitude \:of \:vector\: λa\: is \:  {\bold5\:}}

\:

  • \color{green}{\rm \: a = (−6, 8)}

\:

To Find :

\:  \:

  • \orange{ \rm\: value \:  of \:  \bold{ λ \: }}

\:  \:

\rm \: The \:  magnitude \:  of \:  the  \: vector  \: a  \: is  \:  ||a|| = \sqrt{(-6)^2+ 8^2)}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \:  \:  \rm = \sqrt{( 36+64)} \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:\:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:\:\:   \: \:\:= \rm\sqrt{100} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: \:  \:  \:   \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:\\\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:\: \rm   \underline{\boxed{\red{ \:  = 10 \: }}} \green✓

\:  \:

\rm \: Therefore, \:  λ =   \cancel{\frac{5}{10} } \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \underline  {  \: \boxed{ \red{\: \rm = 0.5\:}}}{  \green ✓ }

Hope helps ~

\color{purple}\frak{SarcasticSmileeᥫ᭡}

`

4 0
2 years ago
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