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2 years ago

PLEASE HELP WITH THE 6 FOLLOWING SCIENCE QUESTIONS (the topic is circuit symbols and equations):

Physics

1 answer:

Citrus2011 [14]2 years ago

6 0

1) I=Q/T then Q= I ×T 10* ( 20*60) = 10 (120) therefore the answer is 1200C

2) I= Q/T that is 500/ 25 and the answer is 20A

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Needing help with this please

Gemiola [76]

Answer:

So A we cant sadly do because we cant draw. B is going to be kinetic. Thats because static friction means it stays in one place, for kinetic it means moving. So it will be 0.05 as the coefficient of the friction. Sadly, I cannot calculate C. You will have to use trigonemetry but I cannot fit that big an explanation.

Answer to A: the free body diagram would be the ski things inclined with gravity, friction, and air resistance. I except you know which directions

Answer to B: Kinetic friction is the answer.

Answer to C: Find on own, I cannot write super big explanations - use trigonometry.

6 0

2 years ago

1.A little girl kicks a soccer ball. It goes 10 feet and comes back to her. How is this possible?

marishachu [46]

Something stopped the force and was able to recreate the same amount of force to send it back to her. Example: A pole

6 0

3 years ago

Read 2 more answers

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

2 years ago

With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how ma

Alenkinab [10]

Answer:

The wavelength stays the same.

Explanation:

When the amplitude is increased, the wavelength stays the same.

Here the wavelength doesn't depend upon the amplitude.

4 0

3 years ago

a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe

Mariana [72]

Answer:

1.34352 kg

Explanation:

$m_w$ = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

$\Delta T$ = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

$m_v$ = Mass of water in the vessel

Here the potential energy will balance the internal energy

$m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg$

Mass of the water in the vessel is 1.34352 kg

6 0

2 years ago