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3 years ago

Please need help with this

Physics

2 answers:

lutik1710 [3]3 years ago

8 0

The first and third choices could both do it, but the first choice makes a much better, clear demonstration.

IgorLugansk [536]3 years ago

4 0

The answer is option A

A current can be passed through the coil with the help of the battery and the presence of magnetic field can be shown by compass.

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A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

Elan Coil [88]

Make the base of the building zero. Then the initial distance is 100m, final distance unknown x. Use gravity, time and initial velocity to solve for final distance.
x - 100 = (0)(5) +(1/2)(-9.81)(5^2)
x - 100 = 0 - 122.625
x = -122.625 + 100
x = -22.625 m below ground

7 0

3 years ago

DESCRIBE THE REQUIREMENTS OF AN INTERNET CONNECTION? please tell me the answer

AlekseyPX

Answer: The basic requirements for connecting to the Internet are a computer device, a working Internet line, and the right modem for that Internet line. In addition, software programs such as Internet browsers, email clients, Usenet clients, and other special applications are needed in order to access the Internet.

Explanation: brainleist pls :)

4 0

3 years ago

7.1 Project Guidelines 2021

Ede4ka [16]

I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and

3 0

3 years ago

A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re

slamgirl [31]

Answer:

tension: 19.3 N
acceleration: 3.36 m/s^2

Explanation:

Given

mass A = 2.0 kg

mass B = 3.0 kg

θ = 40°

Find

The tension in the string

The acceleration of the masses

Solution

Mass A is being pulled down the inclined plane by a force due to gravity of ...

F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

F/m = a = F/m

(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

T = (2(29.4) +3(12.5986))/5 = 19.3192 N

Then the acceleration of B is ...

a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0

3 years ago

A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

$M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})$ where $M_{truck}$ is the mass of the truck, $V_{truck}$ is velocity of the truck, $V_{common}$ is the common velocity of moving and standing truck after collision and $M_{standing}$ is the mass of the standing truck

Making $V_{truck}$ the subject we obtain

$V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}$

Substituting $M_{truck}$ as 25000 Kg, $V_{common}$ as 22.3 m/s, $M_{standing}$ as 2000 Kg we obtain

$V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s$

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0

3 years ago