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2 years ago

Dakdadakdadakdadakda

Physics

1 answer:

notsponge [240]2 years ago

4 0

Answer: dakdadakdadakdadakda

Explanation:(sings) blah blah blah middle fingers in the air l-l-l-loser

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

2 years ago

An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for

Thepotemich [5.8K]

Answer:

21741 s i believe

Explanation:

8 0

3 years ago

A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates 1.5 m/s

Harrizon [31]

We now that follow newton rules f=ma so net force equal to mass*acceleration=>f=50*1.5=75 N

3 0

3 years ago

Read 2 more answers

Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the

Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

$\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}$

The force can be calculated by the formula

$F=k\frac{q1q2}{(r^{\prime})^2}$

Here, k is the constant whose value is

$k=9\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the force will be

$\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}$

7 0

1 year ago

A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci

Simora [160]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

4 0

2 years ago