Diego hypothesizes that the Earth wobbles on its axis in a pattern that is repeated every 20,000 years.

garri49 [273]

Am not really sure but what i see is D

4 0

3 years ago

One hundred turns of insulated copper wire are wrapped into a circular coil of crosssectional area 1.20⇥103 m2. The two ends of

arsen [322]

Answer:

236.3 x $10^-^3$ C

Explanation:

Given:

B(0)=1.60T and B(t)=-1.60T

No. of turns 'N' =100

cross-sectional area 'A'= 1.2 x $10^-^3$ m²

Resistance 'R'= 1.3Ω

According to Faraday's law, the induced emf is given by,

ℰ=-NdΦ/dt

The current given by resistance and induced emf as

I = ℰ/R

I= -NdΦ/dtR

By converting the current to differential form(the time derivative of charge), we get

$\frac{dq}{dt}$ = -NdΦ/dtR

dq= -N dΦ/R

The change in the flux dФ =Ф(t)-Ф(0)

therefore, dq = $\frac{N}{R}$ (Ф(0)-Ф(t))

Also, flux is equal to the magnetic field multiplied with the area of the coil

dq = NA(B(0)-B(t))/R

dq= (100)(1.2 x $10^-^3$ )(1.6+1.6)/1.3

dq= 236.3 x $10^-^3$ C

5 0

3 years ago

A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$