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geniusboy [140]
4 years ago
5

Is eureka wire better conductor than copper?

Physics
1 answer:
Artemon [7]4 years ago
4 0
Copper is a better conductor.
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HELP ASAP!!! The gravitational pull of the Moon is much less than the
katovenus [111]

Answer:

I'm pretty sure its B and C

Explanation:

B bc the weight is gravitational pull x mass so when the object has same mass the weight is smaller on moon

C bc mass is the same - you can't change it

7 0
3 years ago
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A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?
cluponka [151]

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e  -9.8 m/s^2).

The time of the ball in air is 3.2 s, so ascending time is 3.2/2=1.6 s.

Therefore from kinematic equation,

v = u + gt

Substituting the values we get,

0= u - 9.8 (1.6)\\\\u=15.7 m/s, Here v = 0 at top.

Now from equation,

h=ut+\frac{1}{2} gt^2,  here h is the height .

So,

h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m.

Thus, the ball reached at its maximum height of 12.48 m.

8 0
3 years ago
I NEED HELP WITH THE QUESTION ABOVEEE
Sergio039 [100]

I think its b if not then I'm sorry buddy:(

3 0
3 years ago
A ball of mass M and speed of V collides head on with a ball of mass 2M and a speed of v/2, as
pogonyaev

Answer:

2/3V

Explanation:

Given that a ball of mass M and speed V collides with another ball of mass 2M and velocity v/2 . After collision they stick together and we need to find their speed after collision . According to Law of Conservation of Momentum , <em>T</em><em>h</em><em>e</em><em> </em><em>t</em><em>o</em><em>t</em><em>a</em><em>l</em><em> </em><em>m</em><em>o</em><em>m</em><em>e</em><em>n</em><em>t</em><em>u</em><em>m</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>y</em><em>s</em><em>t</em><em>e</em><em>m</em><em> </em><em>b</em><em>e</em><em>f</em><em>o</em><em>r</em><em>e</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>a</em><em>f</em><em>t</em><em>e</em><em>r</em><em> </em><em>c</em><em>o</em><em>l</em><em>l</em><em>i</em><em>s</em><em>i</em><em>o</em><em>n</em><em> </em><em>i</em><em>s</em><em> </em><em>c</em><em>o</em><em>n</em><em>s</em><em>t</em><em>a</em><em>n</em><em>t</em><em> </em><em>.</em><em>T</em><em>h</em><em>a</em><em>t</em><em> </em><em>i</em><em>s</em><em> </em><em>;</em>

\sf\qquad\longrightarrow m_1u_1+m_2u_2=m_1v_1+m_2v_2\\

\sf\qquad\longrightarrow MV + 2M\bigg(\dfrac{V}{2}\bigg) = (2M + M)(v)\\

\sf\qquad\longrightarrow MV +MV= 3Mv\\

\sf\qquad\longrightarrow 2MV = 3Mv\\

\sf\qquad\longrightarrow v =\dfrac{2M}{3M}V\\

\sf\qquad\longrightarrow \pink{ v_{after\ collision}=\dfrac{2}{3}V }

<u>H</u><u>e</u><u>n</u><u>c</u><u>e</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>v</u><u>e</u><u>l</u><u>o</u><u>c</u><u>i</u><u>t</u><u>y</u><u> </u><u>a</u><u>f</u><u>t</u><u>e</u><u>r</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>c</u><u>o</u><u>l</u><u>l</u><u>i</u><u>s</u><u>i</u><u>o</u><u>n</u><u> </u><u>i</u><u>a</u><u> </u><u>2</u><u>/</u><u>3</u><u>V</u><u> </u><u>.</u>

8 0
3 years ago
The total electric flux through a closed cylindrical (length = 1.2 m, diameter = 0.20 m) surface is equal to –5.0 N ⋅ m2/C. Dete
Eddi Din [679]

Answer:

The net charge within the cylinder is (-44.25 pC)

Explanation:

Given that,

Total electric flux, \phi=-5\ N-m^2/C

Length of the cylinder, l = 1.2 m

Diameter of the cylinder, d = 0.2 m              

We know that the Gauss's law of electrostatics gives the relation between electric flux and the net charge. It is given by :

\phi=\dfrac{q}{\epsilon_o}

q is net charge within the cylinder

q=\phi\times \epsilon_o\\\\q=-5\times 8.85\times 10^{-12}\\\\q=-44.25\times 10^{-12}\ C\\\\q=-44.25\ pC

So, the net charge within the cylinder is (-44.25 pC). Hence, this is the required solution.            

6 0
3 years ago
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