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3 years ago

True or False:In a current, electrons will always flow from negative to positive.

2 answers:

7 0

<span>True. Electrons flow from negative toward positive to negative toward positive because opposite charges attract to each other. This is the right answer i took the quiz and this as the answer. Hope it helped.</span>

DIA [1.3K]3 years ago

3 0

It should be true: electrons low from negative toward positive to negative toward positive because opposite charges attract each other.

I hope this was correct

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Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k

NikAS [45]

Answer:

$Fr=12544N$

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation $Fr=usmg$ , where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

$Fr=0.8*1600kg*9.8\frac{m}{s^{2}}$

$Fr=12544N$

5 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago

If 4 x 10^21 electrons pass through a wire connected to a 15 ohm resistor in 3 s, how much electric

sammy [17]

Answer:

1.) Current = 213.33A

2.) Potential difference = 3200 V

Explanation:

Given that

Number of electrons n = 4 x 10^21 electrons

Resistance R = 15 ohm

Time t = 3 s

From the definition of current ;

Current is the rate of flow of changes. That is,

Current I = Q/t

Where

Q = 4 × 10^21 × 1.6 × 10^-19

Q = 460C

Current I = 460/3

Current I = 213.33 A

Using Ohms law which state that

V = IR

Substitute the resistance R and current I into the formula above

Potential difference V = 213.3 × 15

Potential difference V = 3200 V

6 0

3 years ago

An object moves along the x-axis according to the equation x = 3.00t2 – 2.00t + 3.00,

Ray Of Light [21]

Explanation:

x = 3.00 $t^{2}$ – 2.00t + 3.00,

Distance of object at 2 second,

x (t=2) = 3(4) - 2(2) +3

x (t=2) = 12-4 +3

x (t=2) = 11 m

Distance of object at 3 second,

x (t=3) = 3(9) - 2(3) +3

x (t=2) = 27 - 6 + 3

x (t=2) = 24 m

a) the average speed between t = 2.00 s and t = 3.00 s,

Average speed = $\frac{Total distance}{ Total time}$

Average speed = $\frac{x (t=2) + x (t=3)}{3}$

Average speed = $\frac{24+11}{3}$

Average speed = $\frac{35}{3}$

Average speed = 11.66 $\frac{m}{s}$

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = $\frac{dx}{dt}$

Instantaneous speed(v) = 6t - 2 $\left \{ {{t=2} \atop {t=3}} \right.$

Instantaneous speed,v(t=2 to t=3) = 18-2-12+2

Instantaneous speed, v = 6 $\frac{m}{s}$

c) the average acceleration between t = 2.00 s and t = 3.00 s

average acceleration = $\frac{average velocity}{time}$

average acceleration = $\frac{11.66}{3-2}$

average acceleration = 11.66 $\frac{m}{s^{2} }$

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s

instantaneous acceleration = $\frac{dv}{dt}$

instantaneous acceleration =6

instantaneous acceleration = 6 $\frac{m}{s^{2} }$

e) for x =0

0 = 3.00 $t^{2}$ – 2.00t + 3.00

a = 3, b=-2, c=3

t= $\frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}$

t= $\frac{2 \pm \sqrt{4 - 36} }{6}$

t= $\frac{2 \pm \sqrt{-32} }{6}$

general solution of this equation gives imaginary value. Hence, the given object is not at rest.

7 0

3 years ago

Read 2 more answers

Which describes how sliding friction affects pushing a cereal box across a tabletop?

Vinil7 [7]

I think your best bet would be.

It acts in the direction opposite of the motion

5 0

3 years ago

Read 2 more answers