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fenix001 [56]
3 years ago
13

A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the

surface of the earth). What is the direction of the magnetic field?
Physics
1 answer:
Vikki [24]3 years ago
3 0

South

Explanation:

The magnetic force F on a point charge moving with a velocity v in the presence of a magnetic field B is given by

\vec{\textbf{F}} = q\vec{\textbf{v}}\textbf{×}\vec{\textbf{B}}

and according to the right-hand rule, an upward magnetic force on a proton moving westward is only possible if the magnetic field is directed southward.

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What is the net work doneon the object over the distance shown?
GuDViN [60]

A)F_0d

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

so

Area:

\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}

so

\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}

therefore, the answer is

A)F_0d

I hope this helps you

4 0
1 year ago
) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in
Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}  

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0
3 years ago
Imagine that someone is sitting down to enjoy a cup of coffee or hot chocolate. Use your experiences to describe how heat flows
Reika [66]

The heat flows into my body and I start with a warming feeling going down my throat as it starts there I get this warm feeling and I break out in a coldsweat from consuming a warm drink a cold drink will help cool me down :

Explanation:

If drink warm will make u warm drinking cool will make u cool down faster

8 0
3 years ago
Read 2 more answers
Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.
Slav-nsk [51]

Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

4 0
2 years ago
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

4 0
3 years ago
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