Approximately how many calories are in one gram of carbohydrates?

Answer:

6.13 seconds

Explanation:

At the peak of the fireworks trajectory, the velocity of the firework would be zero. Using equation of motion, we have:

v² = u² + 2gh

0 = 40² - (2)(9.81)(h)

0 = 1600 - 19.62h

19.62h = 1600

h = 1600/19.62

h = 81.55 m

Now during the process of explosion, the two parts gained equal vertical momentum but in opposite directions.

We are told the first piece lands in a time of 2.71 s,

Using 3rd equation of motion, we have;

h = ut + ½gt²

81.55 = u(2.71) + ½(9.81 × 2.71²)

81.55 = 2.71u + 36.0228

2.71u = 81.55 - 36.0228

2.71u = 45.5272

u = 45.5272/2.71

u = 16.8 m/s

The time it takes a projectile to return back to its original launch point assuming the projectile was launched

vertically with speed u = 16.8 m/s is;

t = 2u/g

t = (2 × 16.8)/9.81

t = 3.43 s

Thus total time it takes the second mass to reach the ground = 3.43 + 2.71 = 6.13 seconds

Explanation:

In Single Slit Experiment:

The width of the central diffraction maximum is inversely proportional to the width of the slit.

Therefore, if we make the slit width smaller, the angle T(representing the angle between the wave ray to a point on the screen and the normal line between the slit and the screen) increases, giving a wider central band.

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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Answer:

[tex]\Large \boxed{\sf kilo}[/tex]

Explanation:

kilo is a prefix that means [tex]1000[/tex] of the base unit.

Answer:

kilo is the correct answer

Explanation:

because my exam says sooo....

Answer:

Stay the same

Explanation:

Since, friction is negligible:

Initial Momentum = Final Momentum

Initial KE = Final KE

m1 * v1 = m2 * v2

When m increases v decreases.

The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Given:

The paperclips into an open cart rolling along a straight horizontal track with negligible friction,

Calculate the momentum, Since friction is negligible,

Initial Momentum = Final Momentum

Initial Kinetic Energy = Final Kinetic Energy

m₁ × v₁ = m₁  × v₂

When m increases, v decreases,

Thus, momentum will remain the same.

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Answer:

The acceleration of the refrigerator together with the objects decreases.

Explanation:

If the mass of the refrigerator is increased by stacking more masses (objects) on it,

and the force applied remains constant, then we know from

F = ma

where

F is the applied force

m is the total mass of the refrigerator and the objects

a is the acceleration of the masses.

If F is constant, and m is increased, the acceleration will decrease

Answer:

The acceleration decreases.

Explanation:

its right

Answer:

can you tell me about this property

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]

Putting the values

[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

By using lens formula

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

Putting the values

[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]

v = 210 mm .

B ) Magnification is the ratio of the size of the image to the size of the an object.

[tex]\rm magnification = \dfrac{v} { u}[/tex]

[tex]M= \dfrac{210} { 15}[/tex]

M= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

[tex]= 14 \times \dfrac{ 250} { 21}[/tex]

= 166.67

= 170 ( in two significant figures )

Hence all the answers are:

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

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Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

Answer:

-105J

Explanation:

See attached file

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Hope this Helps!! Sorry its late

Answer:

24.34 m/s

Explanation:

recall that one of the equations of motions takes the form:

v = u + at

where,

v = final velocity

u = initial velocity (given as 22.2 m/s)

a = acceleration (given as 2.68m/s²)

t = time elapsed during acceleration (given as 0.80s)

since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:

v = u + at

v = 22.2 + (2.68) (0.8)

v = 24.34 m/s

Answer:

24W

Explanation:

The series connection has a resistance of 2R

The parallel connection has a resistance of R/2 .. the resistance has decreased by a factor 4

Assuming the battery still provides the same pd .. the current increases by a factor of 4 .. increasing the power output by a factor of 4 also (P = V x A)

Power output = 4 x 8W .. .. So P = 24 W

Answer:

energy carried by the current is given by the pointyng vector

Explanation:

The current is defined by

       i = dQ / dt

this is the number of charges per unit area over time.

The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.

But the energy carried by the current is given by the pointyng vector of the electromagnetic wave

            S = 1 / μ₀ EX B

It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement

Answer:

a lower speed

Explanation:

Let us look closely at the Einstein's photoelectric equation;

KE= E-Wo

Where;

KE= kinetic energy of the emitted photoelectron

E= energy of the incident photon

Wo= work function of the metal

Hence,where Wo for metal 1 > Wo for metal 2, it follows that KE for metal 1 must also be less than KE for metal 2.

This is because the difference between E and Wo for metal 1 is smaller than the same difference for metal 2 hence the answer.

Answer:

A) 60 Hz

B) 0.04 T

Explanation:

Given that.

Number of turns, N = 200

Length of the side, l = 20 cm = 0.2 m

Speed if rotation, w = 3600 rpm

Voltage, V = 120 V

First, we try to convert the speed from rpm to rad/s

3600 * (2π/60)

3600 * 0.10473

3600 rpm = 377 rad/s

Now, we use that as our w, speed of rotation

Frequency of output, f =

w/2π

f = 377 / 6.284

f = 59.99 Hz or approximately, 60 Hz.

B

Strength of the magnetic field in which the coil is turning

E• = NABw

Where, A = l² = 0.2² = 0.04, on substituting the values to the equation, we have

120 = 200 * 0.04 * 377 * B

120

Making B subject of formula,

B = 120/ 3016

B = 0.04 T..

The frequency of the output voltage is 60 Hz and the strength of the magnetic field is 0.04 T

Answer:

v/c = 0.76

Explanation:

Formula for Length contraction is given by;

L = L_o(√(1 - (v²/c²))

Where;

L is the length of the object at a moving speed v

L_o is the length of the object at rest

v is the speed of the object

c is speed of light

Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.

Thus;

0.65L_o = L_o[√(1 - (v²/c²))]

Dividing both sides by L_o gives;

0.65 = √(1 - (v²/c²))

Squaring both sides, we have;

0.65² = (1 - (v²/c²))

v²/c² = 1 - 0.65²

v²/c² = 0.5775

Taking square root of both sides gives;

v/c = 0.76

Answer:

As opposed to n-type semiconductors, p-type semiconductors have a larger hole concentration than electron concentration.

Explanation:

In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. A common p-type dopant for silicon is boron or gallium. hope this you :)

Answer:

Here are notes I took on semiconductor conductivity :

________________________________________________________
-A p-type semiconductor is made of a material in which electrical conduction is due to the movement of a positive charge.

-Examples of p-type dopants - boron, aluminum, gallium, indium, and thallium

Explanation:

In this case, indium only correct option being a dopant of a P-type semiconductor. Other options are N-type dopants.

Hopefully its correct !! <3

Answer:

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Explanation:

In this exercise you are asked to relate each with the answers

In general, in the optics diagram,

* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point

* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate

* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.

With these statements, let's review the answers

statement 1 with answer C

statement 2 with answer F

statement 3 with answer B

Statement 1 with E

Statement 2 with A

Statement 3 with D

Answer:

Number of turns of wire(N) = 3,036 turns (Approx)

Explanation:

Given:

Diameter = 13 Cm

emf = 5.6 v

Note:

The given question is incomplete, unknown information is as follow.

Magnetic field increases = 0.25 T in 1.8 (Second)

Find:

Number of turns of wire(N)

Computation:

radius (r) = 13 / 2 = 6.5 cm = 0.065 m

Area = πr²

Area = (22/7)(0.065)(0.065)

Area = 0.013278 m²

So,

emf = (N)(A)(dB / dt)

5.6 = (N)(0.013278)(0.25 / 1.8)

5.6 = (N)(0.013278)(0.1389)

N = 3,036.35899

Number of turns of wire(N) = 3,036 turns (Approx)

Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

[tex]T_1[/tex] = 27°C = (273 + 27) K = 300 K

[tex]T_2[/tex]= 600°C = (600 +273) K = 873 K

speed v = 2 m/s

length x = 40 cm = 0.4 cm

width = 1 cm = 0.001 m

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

[tex]k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg[/tex]

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = [tex]\dfrac{\rho \times v \times x }{\mu}[/tex]

reynolds number = [tex]\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}[/tex]

reynolds number = [tex]\dfrac{0.98}{18.6 \times 10^{-6}}[/tex]

reynolds number = 52688.11204

Prandtl number = [tex]\dfrac{c_p \mu}{k}[/tex]

Prandtl number = [tex]\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}[/tex]

Prandtl number = [tex]\dfrac{0.018693}{0.02476}[/tex]

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate  can be computed as follows:

Nusselt no = [tex]\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}[/tex]

[tex]\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}[/tex]

[tex]\dfrac{h \times 0.4}{0.02476} =161.4252008}[/tex]

[tex]h =\dfrac{161.4252008 \times 0.02476}{ 0.4}[/tex]

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = [tex]h\times A \times (T_2-T_1)[/tex]

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

Answer:

Explanation:

To solve for the power consumed by the trains motor we have to employ the formula for power which is

Power= current * voltage

Given that

voltage V= 800 V

current I= 2130 A

Substituting in the formula for power we have

Power= 2130*800=  1704000 watt

Power = 1704 kW

This is the amount of energy consumed, transferred or converted per unit of time

Hence the power consumed  by the trains motor is 1704 kW

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop.  There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

Answer:

The  value [tex]y = 0.0349 \ m[/tex]

Explanation:

From the question we are told that

   The  distance of the screen is  [tex]D = 2.30 \ m[/tex]

   The  width of the slit is  [tex]d = 0.0368 \ nm = 0.0368 *10^{-3} \ m[/tex]

   The  wavelength is  [tex]\lambda = 558 \ nm = 558 *10^{-9} \ m[/tex]

The  width of the central diffraction peak is  mathematically represented as

        [tex]k = 2 * y[/tex]

Where  y is the distance from the center to the high peak which  is mathematically represented as

       [tex]y = \frac{\lambda * D }{d }[/tex]

substituting values

      [tex]y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }[/tex]

      [tex]y = 0.0349 \ m[/tex]

Answer:

f = 0.4 Hz

Explanation:

The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = radial acceleration = 34.1 m/s²

r = radius = length of beam = 5.55 m

Therefore,

f = (1/2π)√[(34.1 m/s²)/(5.55 m)]

f = 0.4 Hz

Answer:

Explanation:

The center of a lens is known as its optical center. All light rays incident on a particular lens converges at a points a point known as the principal focus or the focal point after reflecting. Note that all light incident on a reflecting surface must all converge at this focal point after reflection.

The distance measured from the center of this lens to its principal focus (otherwise known as focal point) is known as the focal length of the lens.

Based on the explanation above, it cam be concluded that the distance from the center of a lens to the location where parallel rays converge or appear to converge is called the FOCAL length.

Answer:

X and Y are two uncharged metal spheres on insulating stands, and are in contact with each other. A positively charged rod R is brought close to X as shown in Figure (a).

The figure shows two spheres on stands and the positively charged rod. The sphere on the left is marked X. The sphere on the right is marked Y. The spheres are in contact with each other. The rod is marked R and it is located to the left of sphere X.

Sphere  Y  is now moved away from  X , as in Figure (b).

The figure shows two spheres on stands and the positively charged rod. The sphere on the left is marked X. The sphere on the right is marked Y and it is moved away from sphere X. The rod is marked R and it is located to the left of sphere X.

What are the final charge states of X and Y?

Both X and Y are neutral.

X is neutral and Y is positive.

X is positive and Y is neutral.

X is negative and Y is positive.

Both X and Y are negative.

Explanation:

Answer:

E = 2.48 eV

Explanation:

The energy of a photon is given by the following formula:

E = hυ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

υ = frequency of photon = c/λ

Therefore,

E = hc/λ

where,

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5 x 10⁻⁷ m)

E = (3.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 2.48 eV

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

We can calculate the energy (E) of a photon with a wavelength (λ) of 500 nm using the Planck's-Einstein relation.

[tex]E = \frac{h \times c}{\lambda } = \frac{(6.63 \times 10^{-34}J.s ) \times (3.00 \times 10^{8}m/s )}{500 \times 10^{-9}m } = 3.98 \times 10^{-19} J[/tex]

where,

We can convert 3.98 × 10⁻¹⁹ J to eV using the conversion factor 1 J = 6.24 × 10¹⁸ eV.

[tex]3.98 \times 10^{-19} J \times \frac{6.24 \times 10^{18} eV }{1J} = 2.48 eV[/tex]

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

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Answer:

a. 365; b. 3; c. 78; d. 1.343 rad; e. 12; f. 10.66

Explanation:

Assume that the function is

[tex]D(x) = 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12[/tex]

The general formula for a sinusoidal function is

      y = A sin(B(x - C))+ D

   |A| = amplitude

     B = frequency

2π/B = period, P

     C = horizontal shift (phase shift)

     D = vertical shift

By comparing the two formulas, we find

|A| = 3

 B = 2π/365

 C = 78

 D = 12

a. Period

P = 2π/B = 2π/(2π/365) = 2π × 365/2π = 365

The period is 365.

b. Amplitude

|A| = 3

The amplitude is 3.  

c. Horizontal shift

C= 78

The horizontal shift is 78.

d. Phase shift  (φ)

Ths phase shift is the horizontal shift expressed in radians.

φ = C × 2π/365 = 78 × 2π/365 ≈ 1.343

The phase shift is 1.343 rad.

e. Vertical shift

D = 12

The vertical shift is 12.

f. Hours of sunlight on Feb 21

Feb 21 is the 52nd day of the year, so x = 51 (the number of days after Jan 1),

[tex]\begin{array}{rcl}D(x) &=& 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12\\\\&=& 3 \sin (0.01721(51 - 78) ) + 12\\&=& 3\sin(-0.4648) + 12\\&=& 3(-0.4482) + 12\\\&=& -1.345 + 12\\& = & \textbf{10.66 h}\\\end{array}[/tex]

There will be 10.66 h of sunlight on Feb 21 of any given year.

The figure below shows the graph of the function from 0 ≤ x ≤ 365.

Answer:

analyze data without drawing conclusions

Explanation:

Research reports are written in order to communicate clearly, information obtained primarily from research and analysis of data.

Typical reports of scientific research endeavours are written in such a way that they convey the research process succinctly without excessive extraneous information. A report is typically made up of; summary of the contents, introduction/ background, methods, results, discussion, conclusion and recommendations.

Hence a report does not really make inferences from the research findings.

Answer:

When an object move in a straight line without moving back.

Explanation:

Distance is covered by an object is the magnitude of length from one position to the another. It is a scalar quantity.

While displacement is the distance covered in a specific direction. Displacement is a vector quantity. It has both magnitude and direction.

If an object move in a straight path without going back, then, the magnitude of distance will be the same with the magnitude of displacement.

Both distance and displacement are measured in the same unit which is metres.

Therefore, an object have to take a straight path without going back to have the distance and the displacement equal.