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yanalaym [24]
3 years ago
8

How does the

Physics
1 answer:
ASHA 777 [7]3 years ago
3 0

Answer:

The magnetic field strength of an electromagnet is therefore determined by the ampere turns of the coil with the more turns of wire in the coil the greater will be the strength of the magnetic field.

Explanation:

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If the force being applied to an object is doubled, what will happen to its<br> acceleration?
Minchanka [31]

Answer:

Acceleration decreases.

Explanation:

Hope this helps you.

4 0
3 years ago
A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height
nikdorinn [45]

Answer:

Explanation:

According to Equations of Projectile motion :

Time\ of\ Flight = \frac{2vsin(x)}{g}

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

Maximum Height = \frac{(vsinx)^{2} }{2g}

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

Horizontal Range = vcosx * t

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

4 0
4 years ago
The weight of the mass added to the hanger is equal to the extra force on the gas, but what area should we use to calculate the
kari74 [83]

Answer: according to the Avagadro's law, volume is directly propotional to no of moles: VXn

according to the Charles law, volume is directly propotional to  temperatue: VXT

according to the Boyle's law, volume is inversely propotional to P: VX1/P

when we combine them we get:

VXnT1/P

V=knT/P

k= R(universal gas constant)

V=RnT/P

PV=nRT  

8 0
3 years ago
A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee
Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

  • Initial Velocity; u = 0
  • Height from which it has dropped; h = 15m
  • Gravitational field strength; g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2

Final speed of brick as it hits the ground; v =  \ ?

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

v^2 = u^2 + 2gh

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0
2 years ago
Help help help please , i will give brainliest
finlep [7]

Answer:

I believe the answer for #1 is D and the answer for #2 is B

Explanation:

I hope this is correct and helps

6 0
3 years ago
Read 2 more answers
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