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3 years ago

The force on a current-carrying wire is a maximum when the current is moving parallel to a magnetic field. true or false

1 answer:

8 0

The force is given by F=iL×B where i is current, L is the path the current follows (the wire path) and B is the magnetic field. Also recall that the × symbol is the vector cross product, which can be expressed by the sine of the angle between the vectors L and B. Therefore we can also say F=iLBsin(θ). Do you know what value of θ makes sin(θ) the largest? This will tell you if parallel or perpendicular makes the force strongest. For one value of θ it will be zero, and for the other it will be maximized.

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A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following fun

snow_lady [41]

The correct answer would be 1.375 < t < 3 i hope this helps anyone

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3 years ago

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20 POINTS MULTIPLE CHOICE

Shkiper50 [21]

If mass doubled and the force remained constant then the acceleration would also double

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4 years ago

A nautical mile is 6076 feet, and 1 knot is a unit of speed equal to 1 nautical mile/hour. How fast is a boat going 8 knots goin

mezya [45]

Answer:

The speed of the boat is equal to 13.50 ft/s.

Explanation:

given,

1 nautical mile = 6076 ft

1 knot = 1 nautical mile /hour

1 knot = 6076 ft/hr

speed of boat = 8 knots

8 knots = 8 nautical mile /hour

= $8 \times \dfrac{6076\ ft}{1\nautical\ mile}\times \dfrac{1\ hour}{60\times 60\ s}$

= 13.50 ft/s

The speed of the boat is equal to 13.50 ft/s.

5 0

3 years ago

The displacement of a moving object in given interval of time is zero . Would the distance travelled by the object also be zero.

kirill [66]

Hi Pupil Here's Your answer ::

➡➡➡➡➡➡➡➡➡➡➡➡➡

No, on a circle in going from one point around circular path the displacement is zero but the distance covered is equal to the circumference of the circle.

i.e.

$2\pi r$

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this helps . . . . . . .

3 0

3 years ago

A vertical spring with a spring constant of 310 N/m is mounted on the floor. From directly above the spring, which is unstrained

EleoNora [17]

Answer:

7.3cm above the compressed spring.

Explanation:

We can use the conservation energy theorem to solve this problem:

$U_{1}+K_1=U_e+U_2+K_2\\m*g*h+0=\frac{1}{2}k*x^2+0+0\\\\h=\frac{k*x^2}{2*m*g}\\\\h=\frac{310N/m*(3.6*10^{-2})^2}{2*0.28kg*9.8m/s^2}\\\\h=0.073m$

The block was dropped 7.3cm above the compressed spring.

8 0

3 years ago