When placed at a certain point, a

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

Whats the force of a pitching machine on a baseball

ira [324]

Answer:

kinetic

Explanation:

8 0

3 years ago

Read 2 more answers

1. A student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.

Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

$v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h$

Hence, this is the required solution.

8 0

4 years ago

A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a

Fynjy0 [20]

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that elongation due to load given as

$\Delta L=\dfrac{PL}{AE}$

$A=\dfrac{PL}{\Delta LE}$

$A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}$

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

4 0

3 years ago

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The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va

Hitman42 [59]

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

5 0

3 years ago