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belka [17]
3 years ago
6

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Physics
1 answer:
Stolb23 [73]3 years ago
4 0

Answer:

523269.9\ \text{N/m}

Explanation:

q = Charge

r = Distance

q_1=25\ \text{C}

r_1=3000\ \text{m}

q_2=40\ \text{C}

r_2=850\ \text{m}

The electric field is given by

E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}

The electric field at the aircraft is 523269.9\ \text{N/m}

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A dog is running at 15 m/s and then slows down to 10 m/s. It takes him two seconds to do this. What is the dog's acceleration
OLga [1]

Explanation:

a = (Vf-Vi)/t

= (10m/s - 15m/s)/2s

a = -5 / 2 m/s^2

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6 0
3 years ago
Help mee pleaseee :)))
Anettt [7]

Answer:

See the explanation below.

Explanation:

Solving the first image question:

C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.

For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

W = weight [N]

m = mass [kg]

g = gravity acceleration [m/s²]

Now we have

m = 50 [kg]

ge = Earth gravity acceleration = 10 [m/s²]

gp = Distant planet gravity acceleration = 4 [m/s²]

We = ge*m

We = 10*50 = 500 [N]

Wp =gp*m

Wp = 4*50 = 200 [N]

Therefore the answer is D

For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.

m = 70 [kg]

First, we must calculate the acceleration, by means of the following equation of kinematics.

v_{f} =v_{o} +a*t

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 0 (because stars from the rest)

a = acceleration [m/s²]

t = time = 4 [s]

20 = 0 + a*4

20 = 4*a

a = 5 [m/s²]

Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.

F = m*a

where:

F = force [N] (units of Newtons)

m = mass = 2 [kg]

a = acceleration = 5 [m/s²]

F = 2*5

F = 10 [N]

The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]

50 + 40 - 35 - 30 = F

F = 25 [N]

8 0
3 years ago
A figure shows a vertically moving block on the end of a cord. The graph next to the figure gives the vertical velocity componen
Sergio [31]
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3 years ago
A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed
Marina86 [1]

Answer:

a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

Magnetic filed ,B= 0.003 T

Speed ,V= 500 m/s

θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a  ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

a=\dfrac{qVB\ sin\theta}{m}

a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2

a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2

a=0.212 m/s²

4 0
4 years ago
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Answer:

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