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3 years ago

The pressure in a fluid is affected by which characteristic of that fluid?

Physics

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

<h2><em><u>The weight of the fluid</u></em></h2><h2 />

Hope it helps

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A football kick returner catches the ball just as a player from the opposing team dives to tackle him. At the time of impact, th

pochemuha

The total momentum of the players after collision is 130 kgm/s.

The given parameters:

<em>Initial momentum of the returner, </em> $P_i_1$ <em> = 0 kgm/s</em>
<em>The initial momentum of the diving player, </em> $P_i_2$ <em> = 130 kgm/s</em>

The total momentum of the players after collision is determined by applying the principle of conservation of linear momentum as follows;

$P_f = P_i_1 + P_i_2\\\\P_f = 0 + 130\\\\P_f = 130 \ kgm/s$

Thus, the total momentum of the players after collision is 130 kgm/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0

2 years ago

The name used to describe the range of EM waves

Nimfa-mama [501]

Answer: the electromagnetic spectrum

Explanation:

I’m not sure but I think this is it!

8 0

3 years ago

Does anyone taste the difference between left and right twix?

4vir4ik [10]

I personally don't (I can't speak for others tho) but I say right twix is better for the memes

3 0

3 years ago

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

Can any kind soul help me please

Alex

Answer:

I) 420000J

ii)

Explanation:

(I) so you can use the formula for quantity of heat then substitute the values given

formula-Q=mc∆9

3 0

3 years ago