Does changing position of charges change the magnitude

Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien

Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = $\mu *m*g$ .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5 N

F = 86.5 N

so

Acceleration in the First 15 second will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be as

v = u + at ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S in 15 s

s = u × t + 0.5 × a × t²

here u is 0

so distance s will be

s = 0.5 × a × t²

s = 0.5 × 1.15 × 15²

s = 129.37 m

and

now acceleration acting is

F = $\mu *m*g$

m a = $\mu *m*g$

a = $\mu* g$

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s

here v velocity is 0 and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s

solve it we get

s = 151.82 m

so

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

7 0

3 years ago

A moving object is in equilibrium. Which best describes the motion of the object if no forces change?

Alchen [17]

If the object is in equilibrium that means that the sum of the forces on it is zero and the net force is zero. If none of the forces changes then the object continues in constant uniform motion. That means constant speed in a straight line.

8 0

3 years ago

Read 2 more answers

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.320 mm wide. The diffraction pattern is observed

DanielleElmas [232]

Answer:

$W = 10.28\ mm$

Explanation:

Given,

Red light wavelength = 633 nm

width of slit = 0.320 mm

distance,d = 2.60 m

Condition of first maximum

$a sin \theta_1 = m\lambda$

$\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})$

m = 1

$\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})$

$\theta_1 = 0.1133^\circ$

Width of the first minima

$y_1 = L tan \theta_1$

$y_1 = 2.60\times tan( 0.11331)$

$y_1 = 5.14 \ mm$

Now, width of the central region

$W = 2 y_1$

$W = 2\times 5.14$

$W = 10.28\ mm$

8 0

3 years ago

An electric dipole is formed from two charges, ±q, spaced 0.800 cm apart. The dipole is at the origin, oriented along the y-axis

Simora [160]

Answer: q = 2.781e-9C = 2.781nC

E=200C

Explanation:

E = Qd/(2πEor^3)

Where

E=Electric field intensity

Q=Charge

d=distance between the dipole=0.008m

Eo=permitivitty

400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)

Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008

q = 2.781e-9C = 2.781nC

b)

Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:

E = kq*2sin θ/r^2

= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2

= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2

=200 C

3 0

2 years ago

Question 2 (1 point)

sveticcg [70]

according to the facts it should be a numerous amount of wrong answers so ignore dis

7 0

3 years ago