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Misha Larkins [42]
3 years ago
10

Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th

e magnetic force on a honey bee due to Earth's magnetic field as the bee flies near the ground from east to west. Average speed of a bee is 15 miles/hour.
HINT: Model the honey bee as a moving, charged point particle. The direction the bee flies is significant.

FEEDBACK: Model the charged honey bee as a point particle.
Physics
1 answer:
murzikaleks [220]3 years ago
7 0

Answer:

3.35\cdot 10^{-16}N

Explanation:

The force exerted on a charged particle due to a magnetic field is given by:

F=qvB sin \theta

where

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

\theta is the angle between the directions of B and v

In this problem we have:

q=1 pC=1\cdot 10^{-12} C is the charge of one honey bee

v=15 mi/h = 6.7 m/s is the velocity of the bee

B=5.0\cdot 10^{-5} T is the average strength of the Earth's magnetic field

\theta=90^{\circ}, because the bee flies east to west while the Earth's magnetic field is south to north

Substituting into the equation, we find:

F=(1\cdot 10^{-12}C)(6.7 m/s)(5.0\cdot 10^{-5} T)(sin 90^{\circ})=3.35\cdot 10^{-16}N

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A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises
atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s²  * t²

-2.0 m = -4.9 m/s²  * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t      where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t²           y = 1.5 m       y0 = 0 m   t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

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3 years ago
What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k
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We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

\sf \: F=ma

Where,

  • F is force
  • m is mass
  • a is acceleration

In our case,

  • F = ?
  • m = 2500 kg
  • a = 20m/s

\tt \: F_{net}  = 2500 \times 20 \\   \tt= 50000

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>

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Answer:

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Explanation:

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