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3 years ago

Help asapp!!!!!!!!!!!!!!

Physics

1 answer:

IRINA_888 [86]3 years ago

3 0

Sorry don’t know this one

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Several paper clips dangle from the north pole of a magnet. The induced pole in the bottom of the lowermost paper clip is a

sammy [17]

I wanna say it’s B.

4 0

4 years ago

A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the

MaRussiya [10]

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

$v^{2}=u^{2}+2as$ and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

$u=\sqrt {v^{2}-2gs}$

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

$u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s$

5 0

3 years ago

Can someone tell me the answers??? thanks i need it asap!! i will give brainlist!!

Sloan [31]

Increases then decreases

4 0

4 years ago

Read 2 more answers

What is larger a Atom or a Molecule?

Kipish [7]

I believe an Atom is a very powerful source, the basic unit of a chemical element. An atom is a source of nuclear energy.

But a molecule on the other hand isn't so different.
a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
I hope that helps, have a fantastic day!

3 0

3 years ago

Read 2 more answers

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

3 years ago