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Kay [80]
3 years ago
13

An electric field’s direction points from the bottom of the screen (or paper) to the top of the screen. If you place an electron

in the middle of the screen, in which direction will it move?
a. It will move to the top of the screen

b. It will not move

c. It will move to the bottom of the screen

d. It cannot be determined where it will move.
Physics
1 answer:
tekilochka [14]3 years ago
4 0

Answer:

c

Explanation:

Solution) It is option (c)

It will move to bottom of the screen

We know that the direction of electric field is the direction of in which test positive charged is moved. Therefore, electron having negative charged is moved in the direction opposite to electric field.  

Hence, it will move to the bottom of the screen

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A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of
spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as                  

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

                                       Q =  m×c×ΔT

                        Heat gain by aluminium + heat lost by copper  = 0    (1)

For Aluminium:

      Q = m\times0.897\frac{J}{g.k}\times(25-5)

      Q = m x 17.94 joule

For Copper:

Q= 206g\times0.385\frac{J}{g.k} \times(88-25)

       Q= 4996.53 Joule

from eq 1

     m x 17.94 = 4996.53

     mass of aluminium = \frac{4996.53}{17.94} g

    Mass of the aluminium chunk = 278.51 g

                         

3 0
3 years ago
What is the resultant of a pair of one pound forces at right angles to each other?
OleMash [197]

Answer:

F_R=\sqrt{2} \ pound.force

Explanation:

Given that there are two force of 1 pound each at right angles to each other.

The from the vector law of addition:

F_R=\sqrt{F_1^2+F_2^2}

where:

F_R= resultant force

F_1\ \&\ F_2 be the two of the forces to be added.

F_R=\sqrt{1^2+1^2}

F_R=\sqrt{2} \ pound.force

8 0
3 years ago
a car travels between the 100 meter and the 250 meter highway marker in 10 seconds calculate the velocity of the car during this
kozerog [31]

Given:\\\Delta s= 250m-100m=150m\\t=10s\\\\Find:\\v=?\\\\Roz.\\\\v=\frac{\Delta s}{t} \\\\v=\frac{150m}{10s} =15\frac{m}{s}

4 0
3 years ago
An artillery shell of mass 30 kg has a velocity of 250 m/s vertically upward. The shell explodes into two pieces; immediately af
olga_2 [115]

Answer:

9654.34 m

Explanation:

from conservation of momentum

$$\begin{aligned}30 \times 250 &=-10 \times 120+20 \times V \\20 V &=30 \times 250+10 * 120 \\V &=\frac{30 \times 250+10 \times 120}{20}=435 \mathrm{~m} / \mathrm{s}\end{aligned}$$

And from Conservation of Energy

\frac{1}{2} m v^{2}=m g h\\h=\frac{v^{2}}{2 g}\\h=\frac{(435(m/s))^{2}}{2 \times 9.8(m/s^{2} )}\\h=9654.34 (m)

7 0
2 years ago
A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
2 years ago
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