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3 years ago

Which statement correctly describes how a bar magnet should be placed on a globe to correctly align with Earth's magnetic field?

Physics

2 answers:

tatyana61 [14]3 years ago

7 0

Answer: C. Place the magnet vertically on the equator with the north end facing the North pole.

Explanation: A P E X

Dima020 [189]3 years ago

5 0

The answer is B. When the magnet is placed on a globe to correctly align with Earth’s magnetic field, it is considered to be suspended freely. The Earth has geographical poles as well with North and South poles. Since unlike poles attract, the South Pole of the magnet will be attracted to the geographical North.

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A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude

kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

$v_0=115 m/s$ (toward east)

So, decomposing along the x- and y- directions:

$v_{x0} = 115 m/s\\v_{y0} = 0$

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

$a=2.88 m/s^2$ (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

$a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2$

So the two components of the velocity after a time t = 25.0 s will be

$v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s$

So, the magnitude of the velocity of the plane will be

$v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s$

6 0

3 years ago

Calculate the acceleration of a turtle going from 0.3 m/s to 0.7 m/s in 30 seconds.

uysha [10]

Initial velocity=u=0.3m/s
Final velocity=v=0.7m/s
Time=t=30s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{0.7-0.3}{30}$

$\\ \sf\longmapsto Acceleration=\dfrac{0.4}{30}$

$\\ \sf\longmapsto Acceleration=0.01m/s^2$

7 0

3 years ago

When a spring is compressed 2.50 × 10^–2 meter

kompoz [17]

The working of a spring is given by:

$T_{el}= \frac{k\Delta x^2}{2}$

Entering the unknowns, we have:

$T_{el}=\frac{k\Delta x^2}{2} \\ 1.25*10^{-2}*2*J=k(2.5*10^{-2}m)^2 \\ k= \frac{2.5*10^{-2}*J}{(2.5*10^{-2}m)^2} \\ k=40* \frac{J}{m^2} \\ k=40 \frac{N*m}{m^2} \\ \boxed {k=40* \frac{N}{m} }$

If you notice any mistake in my english, please let me know, because i am not native.

4 0

3 years ago

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end o

Misha Larkins [42]

Answer: $P_{bulb}=99.11\ kPa$

Explanation:

Given

When the bulb is squeezed then liquid level rises to height h i.e. pressure decreases inside the bulb which causes rises in the tube

for common elevation i.e. at liquid level pressure must be equal therefore

$P_{bulb}+\rho gh=P_{atm}$

$P_{bulb}=1.013\times 10^5-1490\times 9.8\times h$

for $h=0.15\ m$

$P_{bulb}=1.013\times 10^5-1490\times 9.8\times 0.15$

$P_{bulb}=101.3\times 10^3-2.1903\times 10^3$

$P_{bulb}=99.11\ kPa$

for $h=0.1\ m$

$P_{bulb}=101.3\times 10^3-1.4602\times 10^3$

$P_{bulb}=99.83\ kPa$

8 0

3 years ago

How is the sun rotating

Katyanochek1 [597]

Answer: The Sun rotates on its axis once in about 27 days. ... The Sun's rotation axis is tilted by about 7.25 degrees from the axis of the Earth's orbit so we see more of the Sun's north pole in September of each year and more of its south pole in March.

Explanation: I hope this helps :)

6 0

3 years ago