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2 years ago

Which equation is correct according to Ohm’s law?

Physics

2 answers:

podryga [215]2 years ago

8 0

Where are the options ?

Korvikt [17]2 years ago

4 0

You didn't give us anything other than the question. No options are provided so I cannot answer. Nobody can.

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Two planets have masses 2 x 10^23 kg and 5 x 10^22 kg, and the distance between them is 3 x 10^16 m. Calculate the gravitational

xxTIMURxx [149]

Explanation:

given,

mass of one planet (m1)=2*10^23 kg

mass of another planet (m2)=5*10^22kg

distance between them(d)=3*10^16m

gravitational constant(G)=6.67*10^-11Nm^2kg^-2

gravitational force between them(F)=?

we know,

F=Gm1m2/d^2

or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2

or, F=6.67*2*5*10^-11+23+22/3*3*10^32

or, F=66.7*10^34/9*10^32

or, F=7.41*10^34-32

•°• F=7.41*10^2

thus, the gravitational force between them is 7.14*10^2

3 0

2 years ago

Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

4 0

3 years ago

A 20000 kg railroad car is rolling at 1.00 m/s when a 1000 kg load of gravel is suddenly dropped in. part a what is the car's sp

Talja [164]

Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s

8 0

2 years ago

1. If 800 C of electric charge passed through a light bulb in 4 min; what is the mag

9966 [12]

The electric current passing through the bulb would be 3.3A

<u>Explanation:</u>

Given:

Electric charge, q = 800C

Time, t = 4 min

= 4 X 60 sec

= 240 sec

Electric current, I = ?