Which object had the greatest average speed?
A
B
C
D

Answer:

No, Momentum is conserved.

Explanation:

The ballistic pendulum is a device in which a projectile such as a bullet is fired into a suspended heavy wooden stationary block. Some kinetic energy gets transformed into heat and sound, and some is used to deform the block. However, momentum is conserved.

Answer:

The answer is "[tex]945\ N[/tex]"

Explanation:

Aluminum has 210 megapascals of tensile resistance. They choose a shear force only at bent foil edge to bend aluminum foil over an edge (that is the edge of its box) to pull them. When a roll of aluminium domestic foil is 30 cm in width and about 15.0 micrometers.

[tex]\to 0.000015 \times 0.300 = 0.0000045\\\\\to 210000000 = \frac{F}{0.0000045}\\\\\to F=210000000 \times 0.0000045\\\\\to F = 945\ N[/tex]

Answer:

https://web.njit.edu/~binchen/phys111/LectureNotes/Physics111_lecture08.pdf

Explanation:it's 38 slides so look at the more bottom ones and there will be your question and answer

Answer:

Electricity and magnetism are two related phenomena produced by the electromagnetic force. Together, they form electromagnetism. A moving electric charge generates a magnetic field. A magnetic field induces electric charge movement, producing an electric current.

Answer: x²-x-4

Explanation:

Here, we are going to use the FOIL method to simplify.

--------------------------------------------------------------------------------------

Given

(x+2)(x-1)-2(x+1)

Expand parenthesis

= x²-x+2x-2-2x-2

Put like terms together

= x²-(x-2x+2x)-(2+2)

Combine like terms

= x²-x-4

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Answer:

x² - x - 4

Explanation:

Multiply the parentheses

x² - x + 2x - 2 - 2 (x + 1)

Distribute -2 through the parentheses

- 2 (x + 1)

Since two opposites add up to zero, remove them from the expression

x² - x - 2 - 2

Calculate the difference

x² - x - 4

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec,  B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

a) Determine the angular accelerations of B1 and B2 when couple is applied

at point P ; Co = I* ∝B2'

                150  = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

                  150 = [ (2*0.5^2) / 3  + (2*0.5^2) / 3  + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 =  180 rad/sec

b) Determine the force exerted on B2 at P

T2 = mB1g + T1  -------- ( 1 )

where ; T1 = mB2g  ( at point p )

                 = 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

Answer:

Acceleration = 2.8 m/s²

Explanation:

Given the following data;

Mass = 1500 kg

Net force = 4200 Newton

To find the truck's acceleration;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

[tex] Acceleration = \frac {Net \; force}{mass} [/tex]

Substituting into the formula, we have;

[tex] Acceleration = \frac {4200}{1500} [/tex]

Acceleration = 2.8 m/s²

Answer:

10m/s

Explanation:

velocity=displacement/time

=200/20

=10m/s

Answer:

7

Explanation:

There is really no explanation needed

Answer:

the magnitude of the magnetic field is 1.23 × 10⁻⁸ T  { in the +y direction }

Explanation:

Given the data in the question;

the relation between the electric field and the magnetic field in an electromagnetic wave can be written as follows;

c = E₀/B₀

where c is speed of light, E₀ is the amplitude of the electric field and B₀ is the amplitude of the magnetic field.

given that;  E₀ = 4.70 V/m, we know that; speed of light c = 3 × 10⁸ m/s

so from our relation; c = E₀/B₀

B₀ = E₀/c

we substitute

B₀ = 4.70 / 3 × 10⁸

B₀ = 1.23 × 10⁻⁸ T  { in the +y direction }

Therefore, the magnitude of the magnetic field is 1.23 × 10⁻⁸ T  { in the +y direction }

Answer:

23. Option B. Hertz

29. Refraction.

Explanation:

23. Determination of the unit of measurement of frequency.

Frequency is simply defined as the number of oscillation made in one second. Mathematically, frequency can be represented as:

Frequency = 1/period

f = 1/T

Period is measured in seconds.

Thus, the unit of frequency becomes

f = 1/T

f = 1/s = s¯¹ (Hertz)

Therefore, the unit of frequency is Hertz.

29. When a wave enters a new medium, the speed of the wave is uttered. This leads to the bending of the wave. When this occurs, we can say refraction has taken place.

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

[tex]\lambda = 656.46 \ nm[/tex]

[tex]\Delta \lambda = 0.04[/tex]

[tex]c=3\times 10^8[/tex]

As we know,

⇒  [tex]\frac{\Delta \lambda}{\lambda} =\frac{v}{c}[/tex]

On substituting the values, we get

⇒  [tex]\frac{656.46}{0.04} =\frac{v}{3\times 10^8}[/tex]

⇒         [tex]v=\frac{656.46}{0.04} (3\times 10^8)[/tex]

⇒            [tex]=16411.5\times 3\times 10^8[/tex]

⇒            [tex]=18280 \ m/s[/tex]

or,

⇒            [tex]=18 \ km/s[/tex]

The  speed at which the star is moving relative to the earth is 18 km/s

Since the wavelength is  656.46 nm.

And, Δλ=0.04nm

Here Wavelength should represent the distance between the same points in the cycles of a signal propagated in space or it should be along a wire.

Now the speed should be

[tex]= 0.04 \div 656.46 \times 3 \times 10^8[/tex]

= 1820m/s

= 18 km.s

Therefore, The  speed at which the star is moving relative to the earth is 18 km/s

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Answer:

Planck's radiation law, a mathematical relationship formulated in 1900 by German physicist Max Planck to explain the spectral-energy distribution of radiation emitted by a blackbody (a hypothetical body that completely absorbs all radiant energy falling upon it, reaches some equilibrium temperature, and then reemits

Explanation:

Answer:

C

[tex]6 = \frac{3}{2} \lambda \\ wavelength : \: \lambda = 4 \: m \\ from \: wave \: equation \\ v = f\lambda \\ v = 8 \times 4 \\ v = 24 \: {ms}^{ - 1} [/tex]

The speed of the waves will be =  [tex]24\dfrac{m}{s}[/tex]

Given data is frequency f = 8.0 Hz

As we know that the speed of the wave is given by the formula below

[tex]V=f\times \lambda[/tex]

Since

[tex]6=2 \lambda[/tex]

[tex]\lambda =3m[/tex]

now

[tex]V=8\times3[/tex]

[tex]V=24m[/tex]

The speed of the waves will be = [tex]24\dfrac{m}{s}[/tex]

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Answer:

the length of the column in air

Answer:

c

Explanation:

Answer:

61 N

Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

Volume of brick = 1900 cm³

Density of water = 1 g/cm³

Mass of brick in water =…?

Density = mass / volume

1 = mass of brick in water / 1900

Cross multiply

Mass of brick in water = 1 × 1900

Mass of brick in water = 1900 g

Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

1900 g = 1.9 Kg

Next, we shall determine the weight in water. This can be obtained as follow:

Mass (m) = 1.9 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 1.9 × 10

W = 19 N

Thus, the weight of the brick in water is 19 N.

Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:

Weight in air = 80 N

Weight in water = 19 N

Apparent weight =?

Apparent weight = weight in air – weight in water

Apparent weight = 80 – 19

Apparent weight = 61 N

Answer:

it changes by taking the air from below the plane and curving it to the top causing draw wich slows it down then the weight that pulls it down to land.

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Explanation:

Answer:

351.28 N

Explanation:

Let F be the force on the object and f be the frictional force. The component of the force acting in the horizontal direction causing the object to move is FcosФ where Ф is the angle between F and the horizontal = 43.0°. The frictional force on the packing crate f = μN where μ = coefficient of kinetic friction = 0.35 and N = normal force = W = weight of the packing crate = mg where m = mass of crate = 74.9 kg and g = acceleration due to gravity = 9.8 m/s².  So, f = μN = μW = μmg

So, the net force on the packing crate is

FcosФ - f = ma

FcosФ - μmg = ma

Since the crate moves at constant speed, its acceleration a = 0

So, FcosФ - μmg = ma

FcosФ - μmg = m(0)

FcosФ - μmg = 0

FcosФ = μmg

F = μmg/cosФ

Substituting the values of the variables into the equation, we have

F = μmg/cosФ

F = 0.35 × 74.9 kg × 9.8 m/s²/cos43.0°

F = 256.907 kg-m/s²/0.73135

F = 351.28 kg-m/s²

F = 351.28 N

The continuous changing of rocks from one kind to another is called as the Rock cycle. So, the correct option is D.

The rock cycle is defined as the change of rocks from one form to another as molten magma cools down and becomes an igneous rock which then breaks down into sediments and again forms sedimentary rocks under great pressure. The same rock that can become a metamorphic rock when subjected to great heat and pressure.

Every rock type is altered when it is forced out of its equilibrium conditions. Rock cycle is explained as the chain of processes which produce and transform the rock types in Earth's crust.

Thus, the continuous changing of rocks from one kind to another is called as the Rock cycle. So, the correct option is D.

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A group of organs working together comprises an organ system, B.

Answer:

please give me brainlist and follow

Explanation:

The angle and wavelength at which the light enters a substance and the density of that substance determine how much the light is refracted. ... When light passes from a more dense to a less dense substance, (for example passing from water into air), the light is refracted (or bent) away from the normal.

Answer: b the energy of light...

Explanation:

Answer:

v = 12 m/s

Long, boring, and convoluted explanation:

First, let's lay out our information.

- max height = 3.7 m

- 0 = 45°

- gravitational acceleration constant = 9.8 [tex]\frac{m}{s^2}[/tex]

Since the puma leaves the ground at a 45 ° angle, its motion will follow a curved path as seen in many projectile motion problems, where the object is being influenced solely by the force of gravity. And because the puma leaves the ground at an angle, its initial velocity is broken down into its horizontal and vertical components. We were also told, though indirectly, that the max height is  3.7m  because the puma can reach up to that height. Gravity is always given to be 9.8 [tex]\frac{m}{s^2}[/tex]

Because we are dealing with maximum height and gravity, we have to use the vertical component of the velocity,  vsin ( θ ) , and not the horizontal component, vcos ( θ ) .

Given its max height, the acceleration due to gravity, and the angle, we can now solve for the speed at which the puma leaves the ground using the following equation: vsin ( θ )  = [tex]\sqrt{2hg}[/tex]

Where vsin ( θ )  is the vertical component of the initial velocity and h  and g are max height and gravitational acceleration constant respectively.

Plugin, rearrange and solve

v sin ( θ )  =  [tex]\sqrt{2hg}[/tex]

v sin ( 45 ∘ )  =   √ 2  ×  3.7  ×  9.8

v ( 0.71 )  =  [tex]\sqrt{72.52}[/tex]

v ( 0.71 )  =  8.52

v  =  8.52 /0.71

v =  12 m s

Answer:

[tex]m=3*10^2^3kg[/tex]

Explanation:

From the question we are told that:

Moon diameter [tex]D_m=5480Km \approx 2740000m[/tex]

Earth's frequency [tex]F_e=3.50Hz[/tex]

Planet's frequency [tex]F_p=1.80Hz[/tex]

Generally the equation for Frequency and Gravity relationship is mathematically given by

[tex]\frac{F_p}{F_e}=\sqrt{ \frac{g_p}{g_e} }[/tex]

Therefore  gravity of Moon  is given as

[tex]g_p= g_e*(\frac{f_p}{f_e} )^2[/tex]

[tex]g_p= 9.8*(\frac{1.82}{3.50} )^2[/tex]

[tex]g_p=2.64992m/s^2[/tex]

Generally the equation for mass of spherical body is mathematically given by

[tex]m=\frac{g_pD_m^2}{G}[/tex]

where G= gravitational constant

[tex]m=\frac{2.64992m/s^2*(2740000m)^2}{6.67*10^{-11}}[/tex]

[tex]m=2.98268956*10^2^3kg[/tex]

[tex]m=3*10^2^3kg[/tex]

Answer:

A) determine magnetic fields

For   0 ≤ r ≤ a

Magnetic field = ∅ [tex]\frac{rI}{2\pi a^2}[/tex]

For  a ≤ r ≤ b

Magnetic field =  ∅ [tex]\frac{I}{2\pi r}[/tex]

For   b ≤ r ≤ c

Magnetic field in the region = ∅ [tex]\frac{I}{2\pi r} [ c^2 - r^2 / c^2-b^2 ][/tex]

For  r ≥ c

magnetic filed in the region = 0

B ) attached below

Explanation:

A) Determine the magnetic field in the following regions

i) For   0 ≤ r ≤ a

Magnetic field = ∅ [tex]\frac{rI}{2\pi a^2}[/tex]

attached below is the detailed solution

ii) For  a ≤ r ≤ b

Magnetic field =  ∅ [tex]\frac{I}{2\pi r}[/tex]

attached below is the detailed solution

iii) For   b ≤ r ≤ c

Magnetic field in the region = ∅ [tex]\frac{I}{2\pi r} [ c^2 - r^2 / c^2-b^2 ][/tex]

attached below is the detailed solution

iv) For  r ≥ c

magnetic filed in the region = 0  and this is because the net current enclosed in the region = 0

Answer:

Wavelength = [tex]1.36\times 10^{-34}\ m[/tex]

Explanation:

Given that,

The mass of a ball, m = 0.113 kg

Velocity of the ball, v = 43 m/s

We need to find the wavelength of the ball. It can be calculated by applying the De-Broglie concept. So,

[tex]\lambda=\dfrac{h}{mv}[/tex]

Put all the values,

[tex]\lambda=\dfrac{6.63\times 10^{-34}}{0.113\times 43}\\\\=1.36\times 10^{-34}\ m[/tex]

So, the wavelength of the ball is equal to [tex]1.36\times 10^{-34}\ m[/tex].

Answer:

a) [tex]v=0.999124c[/tex]

b) [tex]E=7.566*10^{22}[/tex]

c) [tex]E_a=760 times\ larger[/tex]

Explanation:

From the question we are told that

Distance to Betelgeuse [tex]d_b=430ly[/tex]

Mass of Rocket [tex]M_r=20000[/tex]

Total Time in years traveled [tex]T_d=36years[/tex]

Total energy used by the United States in the year 2000 [tex]E_{2000}=1.0*10^20[/tex]

Generally the equation of speed of rocket v mathematically given by

[tex]v=\frac{2d}{\triangle t}[/tex]

[tex]v=860ly/ \triangle t[/tex]

where

[tex]\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}[/tex]

[tex]\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}[/tex]

[tex]\triangle t=\sqrt{(860)^2+(36)^2}[/tex]

[tex]\triangle t=860.7532[/tex]

Therefore

[tex]v=\frac{860ly}{ 860.7532}[/tex]

[tex]v=0.999124c[/tex]

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

[tex]E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2[/tex]

[tex]E=7.566*10^{22}[/tex]

c)Generally the equation of the energy [tex]E_a[/tex] required to accelerate the rocket mathematically given by

[tex]E_a=\frac{E}{E_{2000}}[/tex]

[tex]E_a=\frac{7.566*10^{22}}{1.0*10^{20}}[/tex]

[tex]E_a=760 times\ larger[/tex]