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stiv31 [10]
2 years ago
11

Alannah added a strip of zinc to a beaker containing a solution of copper

Physics
2 answers:
USPshnik [31]2 years ago
8 0

The solution of copper sulfate was too cold

Explanation:

enyata [817]2 years ago
3 0

Answer:

too cold

Explanation:

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A periodic wave has a wavelength of 0,50 m and a speed of 20 m/s, What is the wave frequency?
faltersainse [42]

Answer:

u=speed, w=wavelenght, f=frequency

It's known that u=w*f =>  f=u/w

                                       u=20m/s     ==>  f=20/0,5  => f=40 Hz

                                       w=0,50m

Explanation:

6 0
1 year ago
Which best describes the energy of a sound wave as it travels through a medium?
qaws [65]

Answer:

it increases

Explanation:

7 0
2 years ago
Read 2 more answers
Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa
Alex

Answer:

B=1.1636*10^{-3}T

Explanation:

Given data

d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\

To find

Magnitude of the net magnetic field B

Solution

The magnitude of the net magnetic field can be find as:

B=2*u\frac{I}{2\pi r}\\ B=2*(4\pi*10^{-7}  )\frac{32}{2\pi (0.022/2)} \\ B=1.1636*10^{-3}T

3 0
3 years ago
can you help me create a sketch of two different objects with one that has a greater density than the other?
sergey [27]

I don't know how good you are at sketching ... I'm terrible. 
But you can put the point across in a dramatic way if you
can sketch a bowling ball and a basketball ... you'll need
to clearly identify them with the markings you sketch on
each ball. 

They're the same shape and nearly the same size, but
there's a huge difference in their densities.

6 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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