It's a little hard can u help

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Answer 1

Answer: I can't really tell what (I) is, but F) blender def converts electric into motion of spinning blades

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A bug flying horizontally at 1.7 m/s collides and sticks to the end of a uniform stick hanging vertically from its other end. After the impact, the stick swings out to a maximum angle of 7.0Â° from the vertical before rotating back. If the mass of the stick is 16 times that of the bug, calculate the length of the stick (in m).
The unit of electric potential or electromotive force is the
Which describes one feature of the image formed by a convex mirror?????
Inductance is usually denoted by L and is measured in SI units of henries (also written henrys, and abbreviated H), named after Joseph Henry, a contemporary of Michael Faraday. The EMF E produced in a coil with inductance L is, according to Faraday's law, given byE=−LΔIΔt.Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?

Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?

Answers

Answer:

Explanation:

One of the major differences between nuclear reactions and chemical reactions is that nuclear reactions involve larger amount of energy than chemical energy. This is because the force between the protons and neutrons in the nucleus of an atom is much higher than the force of attraction between electrons and the positively charged nucleus, hence nuclear reactions involves/requires a larger amount of energy (because it's reactions involve the nucleus) than chemical reactions (because it's reactions involve the electrons).

Thus, during nuclear fusion, two light nuclei are bombarded against one another to produce a larger/heavier nuclei with the release of large amount of energy (because the forces between the protons and neutrons are much higher) unlike when two atoms/molecules are chemically combined together to form a new molecule with the rearrangement of electrons in the valence shells of the participating molecules.

A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?

Answers

(a) The capacity will be "4006.4 veh/h".

(b) The speed at capacity be "25 mph".

(c) The density will be "299 veh/mi".

Given:

$q = 50k - 0.156 k^2$

At max. flow density,

$(dd)/(dk) =0$

$((dq)/(dt) ) = 50-0.321k =0$

(a)

→ $k = ((50)/(0.312) )$

$= 160.3 \ or \ 160 \ veh/mi$

By substituting the value,

→ $q = 50k-0.156k^2$

$= 50* 160.3-0.156* (160.3)^2$

$= 4006.4 \ veh/h$

(b)

The speed will be:

→ $U = (q)/(k)$

$= (4006.4)/(160.3)$

$= 25 \ mph$

(c)

The density be:

→ $1001.6 = 50k-0.156k^2$

$0.156k^2-50k+1001.6 =0$

$k = 21.5 \ veh/mi \ or \ 299 \ veh/mi$

Thus the responses above are correct.

Find out more information about density here:

brainly.com/question/6838128

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region

Answers

Answer:

$2.429783984* 10^(-14)\ A$

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = $1.6* 10^(-19)\ C$

Number of electrons passing per second

$n_e=(6020)/(0.0476)\n\Rightarrow n_e=126470.588$

Number of protons passing per second

$n_p=(1681)/(0.0662)\n\Rightarrow n_p=25392.749$

Current due to electrons

$I_e=n_ee\n\Rightarrow I_e=126470.588* 1.6* 10^(-19)\n\Rightarrow I_e=2.0235* 10^(-14)\ A$

Current due to protons

$I_p=n_pe\n\Rightarrow I_p=25392.749* 1.6* 10^(-19)\n\Rightarrow I_p=4.06283984* 10^(-15)\ A$

Total current

$I=2.0235* 10^(-14)+4.06283984* 10^(-15)\n\Rightarrow I=2.429783984* 10^(-14)\ A$

The average current is $2.429783984* 10^(-14)\ A$

A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

Answers

Answer:

The acceleration of the mower will be "4.7 m/s²".

Explanation:

Balance of vertical force will be:

⇒ $Ra + Rb = mg$

For wheel to take off at A,

⇒ $Ra = 0$

Hence,

$Rb=mg$

Balancing moments about G will be:

⇒ $F* h = Rb* LB$

As we know,

Force, F = $(Rb* LB )/(h)$

On putting the values, we get

⇒ = $(38* 9.81* 0.36)/(0.75)$

⇒ = $178.9 \ N$

Now,

Acceleration, a = $(F)/(m)$

⇒ = $(178.9)/(38)$

⇒ = $4.7 \ m/s^2$

Determine how many wavelengths will fit into the glass container when it is a vacuum. Since the light passes through the container twice, you need to determine how many wavelengths will fit into a glass container that has a length of 2L.

Answers

Answer:

# = 2L /λ

Explanation:

For this exercise we can use a direct proportion rule. If there is 1 wave in 1 wavelength in 2L wave how many lengths are there

# = 2L /λ 1 wave

let's calculate

# = 2L /λ

we see that the longer the wavelength the fewer waves fit in the container

Four students measured the same line with a ruler like the one shown below. The results were as follows: 5.52 cm, 6.63 cm, 5.5, and 5.93. Even though you cannot see the line they actually measured, which of the recorded measurements are possible valid measurements for this instrument, according to its precision? Select all that apply. 1. 5.52 2. 6.63 3. 5.5 4. 5.93

Answers

Answer:

correct answer is 1 and 3

Explanation:

In direct measurement with an instrument, the precision or absolute error of the instrument is given by its appreciation, in this case we see that the measurements have two decimal places, so the appreciation of the instrument must be 0.01 cm

Based on this appreciation, the valid measurements are 5.52 and 5.5.

the other two measurements have errors much higher than the assessment of the instrument, for which there must have been some errors in the measurement.

The correct answer is 1 and 3

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Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement?(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?(c) What was his total displacement relative to the air mass?