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nataly862011 [7]
3 years ago
5

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

o say that the lines we use to represent an electric field indicate the direction in which a _______ test charge would initially move when released from rest. Which of the following fills in the three missing words correctly?
a. (positive; negative; negative)
b. (positive; negative; positive)
c. (negative; positive; negative)d. (negative; positive; positive)
Physics
1 answer:
lesantik [10]3 years ago
3 0

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would  be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

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Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k
Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is \bf{6.59~m/s}.

Part(B): The direction is \bf{42.7^{0}} south to east.

Explanation:

Given:

Mass of Jack, m_{1} = 59.0~Kg

Mass of Jill, m_{2} = 47..0~Kg

Initial velocity of Jack, v_{1i} = 8.00~m/s

Initial velocity of Jill, v_{2i} = 0

Final velocity of Jack, v_{1f}  5.00~m/s

The final angle made by Jack after collision, \alpha = 34.0^{0}

Consider that the final velocity of Jill be v_{2f} and it makes an angle of \beta with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}

where, v_{2f}^{x} is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s

Conservation of momentum of the system along north direction is given by

~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s

where, v_{2f}^{y} is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s

Part(B):

The direction is given by

\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}

4 0
4 years ago
A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01x10⁵ Pa.
Cerrena [4.2K]

Answer:

\frac{1.019}{1}

Explanation:

To solve this equation we will have to consider that the bubble is filled with an Ideal Gas and as such we can use the Ideal Gas Law

PV = nRT

Where

P = Pressure

V = Volume

n = Moles

R = Ideal Gas Constant

T  = Temperature

Now since we know that the value for the temperature and moles is constant we can simply use Boyles Law for the two states

P_{1} V_{1} =P_{2} V_{2}

Let us look at the two states

State 1 (at top)

Pressure = 1.01*10^5

Volume = V_{1}

State 2 (at bottom)

Pressure = 1.01*10^5 + dgh

Where

d = Density of liquid (1000 kg/m³)

d = Acceleration due to gravity (9.8 m/s²)

d = Height of liquid (0.200 m)

Pressure = 102,962

Volume = V_{2}

Inputting these values into the Boyles Law

P_{1} V_{1} =P_{2} V_{2}\\ (101000)V_{1} = (102962)V_{2}\\ \frac{V_{1}}{V_{2}} = \frac{102962}{101000} \\  \frac{V_{1}}{V_{2}} = \frac{1.019}{1}

6 0
3 years ago
John’s car travels a distance from A to B at a speed of 50 km/h for two hours as from B to C at a speed of 40 km/h for two hours
Blababa [14]

Please find attached photograph for your answer. Hope it helps. Please do comment.

7 0
3 years ago
If the radius of an object in circular motion is doubled, what change will occur in the centripetal force?
cestrela7 [59]

Answer:

The centripetal force will be 1/2 as big as it was. (option c)

Explanation:

Recall that centripetal force (F_c) is defined as: F_c=m\,* \frac{v^2}{r} where "v" is the tangential velocity of the object in circular motion, "r" is the radius of rotation and "m" is the object's mass.

So if we start with such formula with a given mass, radius, and tangential velocity, and then we move to a situation where everything stays the same except for the radius which doubles, then the new centripetal force (F'_c) will be given by: F'_c=m\,* \frac{v^2}{2r}

and this is half (1/2) of the original force:

F'_c=m\,* \frac{v^2}{2r}\\F'_c=m\,* \frac{v^2}{r}*\frac{1}{2} \\F'_c=F_c\,*\,\frac{1}{2}

which is expressed by option "c" of the provided list.

8 0
3 years ago
Read 2 more answers
Why does a hot air balloon float?
GaryK [48]
The most accurate answer among the choices would be the fourth one, because if the weight of the air displaced is greater than the balloon's weight, the balloon will float upwards. Density of the air also plays a part. Hot air = less dense. Hope my answer has come to your help.
8 0
3 years ago
Read 2 more answers
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