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3 years ago

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A copper bar of length h and electric resistance r slides with negligible friction on metal rails that have negligible electric

resistance (see figure below. the rails are connected on the right with a wire of negligible electric resistance

1 answer:

Elanso [62]3 years ago

3 0

I think math is the best idea

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A 6.00 A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5×1028 free elect

navik [9.2K]

Answer:

Explanation:

Current, I = 6 A

diameter of wire, d = 2.05 mm

number of electrons per unit volume, n = 8.5 x 10^28

If the diameter is doubled,

The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.

8 0

2 years ago

What are the comparisons and contrast of adaptation and natural selection? ASAP!!!!!!

Andre45 [30]

Answer:

Natural selection is a mechanism, or cause, of evolution.

Explanation:

Adaptations are physical or behavioral traits that make an organism better suited to its environment. Heritable variation comes from random mutations.

5 0

2 years ago

a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²

tigry1 [53]

At the player's maximum height, their velocity is 0. Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

which tells us the player's initial velocity $v_i$ is

$0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}$

The player's height at time $t$ is given by

$y=v_it-\dfrac g2t^2$

so we find their airtime to be

$0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s$

6 0

3 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

3 years ago

Let’s tie this all together. It makes sense that, if the rope force remains greater than the gravitational force, the child keep

Pavlova-9 [17]

Answer:

If child weight is equal to rope force then child will move with uniform speed

or we can say that the child will remain at rest in his position

Explanation:

As we know that child is hanging by rope

so here there will be two forces on the child

1) Weight or gravitational force which act vertically downwards

2) Tension in the rope which act vertically upwards

Now if child will accelerate upwards then tension force must be more than the weight of the child

If tension force is less than the weight then child will decelerate and his speed will decrease

if tension force is equal to child weight then in that case the child will remain at rest or it will move with same speed

8 0

2 years ago