An electric current is flowing through a cord what will happen if a magnet is brought near the cord

What additions must you make to the cell for it to generate a standard emf?

xeze [42]

Answer: Addition of salt bridge

To generate a standard electromotive force, an addition of salt bridge is essential to help maintain electrical neutrality within the internal circuit and prevent the cell from rapidly running its reaction to equilibrium.
In addition, salt bridge contains potassium sulfate, an inert electrolyte where its ions will diffuse into the separate half-cells to stabilize the building charges at the electrodes and produce electricity.

4 0

4 years ago

A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will

gtnhenbr [62]

1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
$x(t)=v_0t$
$y(t)=h- \frac{1}{2}gt^2$
where the motion on the x-axis is a uniform motion with constant speed $v_0=18.1 m/s$ , while the motion on the y-axis is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ , and initial height $h=1.85 m$ .

First of all, we can find the time t at which the ball reaches the ground by requiring $y(t)=0$ :
$0=h- \frac{1}{2}gt^2$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }= 0.61 s$

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
$x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m$

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
$v_x=18.1 m/s$
On the y-axis, the velocity is given by
$v_y(t)=gt$
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
$v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s$
And the speed of the ball is the magnitude of the resultant of the two components:
$v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s$

8 0

3 years ago

Read 2 more answers

Question 4 of 10

ch4aika [34]

Habitat fragmentation is a cost of urban development.

Option: A

Explanation:

Though from the view point or perspective of up gradation and development urban development is much needed but in cost of habitat fragmentation which feels very bitter. As habitat fragmentation leads to the loss of habitat, disruption of ecological cycle and environmental equilibrium.

Actually in the name of urban development we the human use our bread giver environment in a wrong way which causes natural disasters in long run. Animals become endangered , vulnerable and extinct with passage of time. Because they forced to enter into human settlements.

4 0

3 years ago

Draw the graphs for exothermic and endothermic reactions Label: axes, reactants, products, Energy of reaction, heat energy chang

nikdorinn [45]

Answer:

axes, reactants, products, Energy of reaction, heat energy

Explanation:

axes, reactants, products, Energy of reaction, heat energyaxes, reactants, products, Energy of reaction,grreactantsaphs heat energy

3 0

3 years ago

A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train th

Rasek [7]

Answer:

0.1575 m/s^2

Explanation:

Solution:-

- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).

- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.

- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).

- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:

$v_f^2 = v_i^2 + 2*a*( s_f - s_o )$

- We will plug in the given parameters in the equation of motion given above:

$42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}$

Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2

7 0

3 years ago