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4 years ago

Benny has 20 jellybeans and wants to share with his friends how many will each friend get? there are 5 friends.

1 answer:

7 0

Answer each friend will get 3.33333 repeating if he is included. if only his friends are getting them then each one gets 4

Explanation:

devide 20/6 and 20/5 respectively.

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You are approaching an intersection where you want to make a turn, and you have a green

Nat2105 [25]

Slow down to a safe turning speed before entering the intersection

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3 years ago

Where is the distribution of positive charges found in the atom?

Agata [3.3K]

I think the answer is proton

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4 years ago

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. How much power is this?

Alex_Xolod [135]

A.900 watts That would be your correct answer

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3 years ago

Read 2 more answers

A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before co

alina1380 [7]

Answer:

The force exerted by the wood on the bullet is 399.01 N

Explanation:

Given;

mass of bullet, m = 0.0021 kg

initial velocity of the bullet, u = 497 m/s

final velocity of the bullet, v = 0

distance traveled by the bullet, S = 0.65 m

Determine the acceleration of the bullet which is the deceleration.

Apply kinematic equation;

V² = U² + 2aS

0 = 497² - (2 x 0.65)a

0 = 247009 - 1.3a

1.3a = 247009

a = 247009 / 1.3

a = 190006.92 m/s²

Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;

F = ma

F = 0.0021 x 190006.92

F = 399.01 N

Therefore, the force exerted by the wood on the bullet is 399.01 N

6 0

3 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

$\epsilon_{max} = NBA\omega$

Where,

N= Number of turns of the coil

B = Magnetic field

$\omega =$ Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

$\omega = 2\pi f$

$\omega = 2\pi*61.5$

$\omega =123\pi rad/s$

Replacing at the equation our values given we have that

$\epsilon_{max} = NBA\omega$

$\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega$

$\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)$

$\epsilon_{max} = 1.5732*10^{-11}V$

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

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4 years ago