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uysha [10]
3 years ago
10

PLEASE HELP ME WITH THIS ONE QUESTION

Physics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

B) The current of the entire circuit is lowered.

Explanation:

There are two ways in which a resistor can be connected in a circuit;

i) Series: In a series connection, resistors are connected end to end. In this arrangement, the current passing through each resistor is the same while the potential difference across each resistor is different.

ii) Parallel: In a parallel connection, resistors are connected to  common junctions. In this arrangement, the potential difference across each resistor is the same while the current across each resistor varies.

In a series circuit, the effective resistance is the algebraic sum of resistance of all the resistors connected in series. Hence, if a large resistor is connected in series, the effective resistance of the circuit is greatly increased.

Since the voltage is given by;

V=IR (Ohm's law)

Where R is the effective resistance

Then;

I = V/R

If R becomes very large due to a large resistance connected in series, then the current of the entire circuit is lowered.

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Relation between wavelength wave velocity and time period
Snezhnost [94]

Answer: wavelength=velocity×period

Explanation:the relation between velocity, wavelength and period is

Wavelength=velocity×period

8 0
2 years ago
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian
Llana [10]

Complete question

The complete question is shown on the first uploaded image  

Answer:

The velocity is  v = c* \sqrt{1 -  \frac{1}{n^2} }

Explanation:

From the question we are told that

           a = nb

The length of the minor axis  of  the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the  Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

        h = t \sqrt{1 - \frac{v^2}{c^2} }

Here t  is the actual length of the major axis of of the  Empire's symbol, (an ellipse)

So t = a = nb

and  b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the  Empire's symbol, (an ellipse)

 i.e    h = b

So

    b  =  nb  [\sqrt{1 - \frac{v^2}{c^2} } ]  

     [\frac{1}{n} ]^2 =  1 -  \frac{v^2}{c^2}

      v^2 =c^2 [1- \frac{1}{n^2} ]

       v^2 =c^2 [\frac{n^2 -1}{n^2} ]

        v = c* \sqrt{1 -  \frac{1}{n^2} }

     

     

6 0
3 years ago
A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the
Ilya [14]

Answer:

d) is the same as when it started from rest

Explanation:

using equation of motion

v = u + at

second law of momentum defines

F = ma

a = F /m

the equation becomes

v = u + (F/m)t

from hear

since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.

5 0
3 years ago
How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
Anestetic [448]

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

d=vt

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

d=(2.1)(10)=21 m

3 0
3 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit
creativ13 [48]

Answer:

Explanation:

From the given information:

Distance d_i = 4.8 \times 10^{10} \ m

Speed of the comet V_i = 9.1 \times 10^{4} \ m/s

At distance d_2 = 6 \times 10^{12} \ m

where;

mass of the sun = 1.98 \times 10^{30}

G = 6.67 \times 10^{-11}

To find the speed V_f:

Using the formula:

E_f = E_i + W \\ \\  where; \  \  W = 0  \ \  \text{since work done by surrounding is zero (0)}

E_f = E_i + 0 \\ \\  K_f + U_f = K_i + U_i  \\ \\ = \dfrac{1}{2}mV_f^2 +  \dfrac{-GMm}{d^2} =  \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [  \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}

V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [  \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}

\mathbf{V_f =53.125 \times 10^4 \ m/s}

3 0
2 years ago
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