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LiRa [457]
3 years ago
9

Which of the following is an x-intercept of the function below

Physics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

Intercepts occur at    x = 1, y = 0

                                   x = 3, y = 0

Only (D) is correct

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An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases
egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

7 0
3 years ago
A Net Force of 9.0 N acts through a distance of 3.0 m in a time of 3.0 s. The work done is?
JulijaS [17]
Work done is the distance a force acts over.

So, the work done here is 9.0N * 3.0m = 27 J
5 0
3 years ago
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Answer these multiple choice questions!!! 20 points!!!
irga5000 [103]
I need pictures or something
7 0
3 years ago
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Which substance is a heterogeneous mixture?
anyanavicka [17]

Answer:

A heterogeneous mixture is simply any mixture that is not uniform in composition - it's a non-uniform mixture of smaller constituent parts.

Explanation:

3 0
3 years ago
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Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affec
lara31 [8.8K]

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1

Therefore, the force between the charges are not affected.

7 0
3 years ago
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