A 72-tooth gear is connected to a 12-tooth gear. If the large gear makes one complete turn, how many turns

Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric ch

Eduardwww [97]

Answer:

a) True

Explanation:

There are several possible explanations for this positive charge

* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,

to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct

* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.

When reviewing the different statements we have

a) True, it agrees with the second explanation of the phenomenon

b) False. The earth is a deposit of negative charge

c) false. If this is the case the charge should be negative

d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.

6 0

3 years ago

Tarzan, who weighs 825 N, swings from a cliff at the end of a 19.7 m vine that hangs from a high tree limb and initially makes a

kodGreya [7K]

Answer:

a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,

c) F = 281.70 N, d) θ = 338.58º , e) a = 3,588 m / s² , f) θ = 201.45º

Explanation:

For this exercise we will use Newton's second law on each axis

X axis

-Tₓ = m aₓ

Y Axisy

$T_{y}$ –W = m a_{y}

Let's use trigonometry to find the components of force

sin 21.5 = Tₓ / T

cos 21.5 = T_{y} / T

Tₓ = T sin 21.5

T_{y} = T cos 21.5

Tₓ = 768 sin 21.5 = 281.47 N

T_{y} = 768 cos 21.5 = 714.56 N

a) the force of the rope on Tarzan is

T = (281.47 i ^ + 714.56 j ^) N

b) The net force is the subtraction of the tension minus the weight of Tarzan

Y Axis F_net = 714.56 - 825 = -110.44 N

F_net = (281.47 i ^ - 110.44 j ^) N

c) Let's use Pythagoras' theorem

F = √ (Fₓ² + T_{y}²)

F = √ (281.47² + 110.44²)

F = 281.70 N

d) Let's use trigonometry

tan θ = F_{y} / Fₓ

θ = tan⁻¹ F_{y} / Fₓ

θ = tan⁻¹ (-110.44 / 281.47)

θ = -21.42º

This angle is average clockwise, for counterclockwise measurement

θ = 360 - 21.42

θ = 338.58º

Acceleration

X axis

Tₓ = m aₓ

aₓ = Tₓ / m

The mass of Tarzan is

m = W / g

m = 825 / 9.8 = 84.18 kg

aₓ = 281.47 / 84.18

aₓ = -3.34 m / s2

Y Axis

T_{y}-W = m a_{y}

a_{y} = (T_{y} -W) / m

a_{y} = (714.56-825) / 84.18

a_{y} = - 1,312 m / s²

Acceleration Module

a = √ aₓ² + a_{y}²

a = √ (3.34² +1.312²)

a = 3,588 m / s²

The angle

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-1312 / -3.34)

θ = 21.45º

Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis

θ = 180 + 21.45

θ = 201.45º

3 0

3 years ago

A long, thin solenoid has 390 turns per meter and a radius of 1.20 cm. The current in the solenoid is increasing at a uniform ra

gtnhenbr [62]

To solve this problem it is necessary to apply the concepts related to Faraday's law and the induced emf.

By definition the induced electromotive force is defined as

$\int E dl = -\frac{d\phi}{dt}$

$\int E dl = -(\frac{dB}{dt})A$

Where,

$\phi =$ Electric field

B = Magnetic Field

A = Area

At the theory the magnetic field is defined as,

$B = \mu_0 NI$

Where,

N = Number of loops

I = current

$\mu_0 =$ Permeability constant

We know also that the cross sectional area, is the area from a circle, and the length is equal to the perimeter then

A = \pi r^2

l = 2\pi r

Replacing at the previous equation we have that

$E (2\pi r) = \mu_0 n (\frac{di}{dt})(\pi R^2)$

Where,

R = Radius of the solenoid

r = The distance from the axis

Re-arrange to find the current in function of time,

$\frac{di}{dt} = \frac{Er}{\mu_0 NR^2}$

Replacing our values we have

$\frac{di}{dt} = \frac{(8.00*10^{-6})(0.0348)}{(4\pi*10^{-7})(390)(1.2*10^-2)^2}$

$\frac{di}{dt} = 3.94487A/s$

8 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

adoni [48]

Answer:

B) R1 = 6 V and R2 = 6V

Explanation:

In series, both resistors will carry the same current.

that current will be I = V/R = 12 / (10 + 10) = 0.6 A

The voltage drop across each resistor is V = IR = 0.6(10) = 6 V

6 0

3 years ago

Pleaseeeee help me with b, c, and d. There are no angles.

taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, $F_{NET}$ = F - $F_{Friction}$ = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

$F_{Friction}$ = The force of friction

∴ $F_{NET}$ = m × a = m × 0 m/s² = 0 N

Therefore;

$F_{NET}$ = F - $F_{Friction}$ = 0 N

From which we have;

F = $F_{Friction}$ = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = $F_{Friction}$ × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, $d_p$ = 0 m

∴ The work done by the normal direction = N × $d_p$ = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, $d_{wieght}$ = 0 m

Therefore, the work done by the weight = W × $d_{wieght}$ = W × 0 m = 0 J

The work done by the weight = 0 J

7 0

3 years ago