(a) The velocity is "6.2 m/s".
(b) The average force is "800.83 N".
According to the question,
Force,
F = 800 NDeflection,
x = 1.2 cm= [tex]1.2\times 10^{-2} \ m[/tex]
As we know,
The work done,
→ [tex]W = F\times d[/tex]
[tex]= 800\times 1.2\times 10^{-2}[/tex]
[tex]= 9.6 \ J[/tex]
(a)
Given:
Mass of hand,
m = 0.50 kgNow,
→ [tex]\frac{1}{2} mv^2 = 9.6 \ J[/tex]
[tex]v = \sqrt{\frac{2\times 9.6}{0.50} }[/tex]
[tex]= 6.2 \ m/s[/tex]
(b)
→ [tex]v^2 = u^2 +2ax[/tex]
→ [tex]a= \frac{v^2}{2x}[/tex]
[tex]= \frac{(6.2)^2}{2\times 1.2\times 10^{-2}}[/tex]
[tex]= 1601.67 \ m/s^2[/tex]
hence,
The average force will be:
→ [tex]F_{avg} = m\times a[/tex]
[tex]= 0.50\times 1601.67[/tex]
[tex]= 800.83 \ N[/tex]
Thus the above answers are correct.
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1. What is the relationship between electricity and magnetism?
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.
How does density affect refraction
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.
A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1.00 g cm³?
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
How did the projected storm track change from image A TO B TO C TO D
Please help 23 also 29 the answer choices are reflection or refraction your fraction absorption
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.
what part of the rocket reaches space
Answer:
7
Explanation:
There is really no explanation needed
A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. Part A If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 4.70 V/mV/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
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 }
simplify. (x+2)(x-1)-2(x+1)
Answer: x²-x-4
Explanation:
Here, we are going to use the FOIL method to simplify.
FirstOuter Inner Last--------------------------------------------------------------------------------------
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
Hope this helps!! :)
Please let me know if you have any questions
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
Groups of organs that work together to complete a series of takes are called:
A: Skeletal System
B: Organ System
C: Muscular System
A group of organs working together comprises an organ system, B.
The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (This arrangement is called a double pendulum.) The counterclockwise couple C0, having moment 150????⋅m, is applied to B2beginning at t=0. Find the angular accelerations of B1and B2upon application of the couple, and the force exerted on B2at P.
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
PLEASE HELP! URGENT
Explain how the forces need to change so an aeroplane can land
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.
please put me brainliest
Explanation:
Discuss how planck's hypothesis explain the observed blackbody spectrum.
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:
- Is the kinetic energy conserved in the collision between
the ball and the pendulum?
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.
3.9. How long will a boy sitting near the window of a train travelling at 36km/h see a train passing by in the opposite direction with a speed of 18km/h. The length of the slow-moving train is 90 m.
Answer:
please give me brainlist and follow
Explanation:
Given:
The speed of the train ,which the boy is observing(t) = 18 km/hr
[ Now , to convert km/hr into m/s we have to multiply by 5/18 ]
So,
speed of the observed train = 18× 5/18
☞ 05 m/s
Also, The speed of the train in which the boy is sitting is(s) = 36 km/hr
Speed in m/s will be = 36 × 5/18
☞ 10 m/s
Also, The Length of the Train = 90m
Therefore:.
The speed of the train is given (expressed) by
= > speed \: = \frac{distance}{time}=>speed=
time
distance
= > time \: = \frac{distance}{Velocity }=>time=
Velocity
distance
Where,
velocity = Realtive velocity of both trains
Distance = Length of the train
So, The above mentioned formulae can also be written as :
= > time = \frac{length}{relative \: velocity}=>time=
relativevelocity
length
Here,
Relative velocity will be :
☞. s -(-t)
☞. 10 -(-5)
☞.15 m/s = Relative velocity
Now , Finally putting the values in the formulae respectively
■ Time = 90/15
■ Time = 6 sec [Answer]
A 1,500-kg truck has a net force of 4,200 N acting on it. What is the truck’s acceleration?
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
Aluminum has a shear strength of 210 megapascals. When you bend aluminum foil around an edge (i.e., the edge of the box) and pull, you are effectively applying a shear force along the bent edge of the foil. If a roll of household aluminum foil is 30.0 centimeters wide and its thickness is approximately 15.0 micrometers, how much shear force is needed to pull off a sheet
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]
The continuous changing of rocks from one kind to another is called:
Question 5 options:
Metamorphosis
Extrusion
The carbon cycle
The rock cycle
The continuous changing of rocks from one kind to another is called as the Rock cycle. So, the correct option is D.
What is Rock cycle?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 capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d
a. True
b. False
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
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].
What is the speed of a 1.57 N bullet traveling with a total momentum equal to 131 Kg.m/s?
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
A system consists of two particles. Which of the following scenarios would the force on each particle increase the most? If multiple scenarios would increase it by the same amount, select all that apply.
Group of answer choices
Doubling the distance between the particles.
Halving the distance between the particles.
Doubling one charge.
Doubling both charges
Doubling one charge, halving the other.
Halving one charge.
halving both charges.
Answer:
c
Explanation:
The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7m when leaving the ground at an angle of 45 degrees. with what speed must the animal leave the ground to reach that height?
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
A graph of a wave is shown if the frequency f is 8.0 Hz what is the waves speed
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
What will be the speed of the wave?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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When measuring speed of sound in air, speed of light = 4f(length of column of air + 0.3 * diameter of tube).
Why is multiplied by 4?
Where does the 4 come from?
Answer:
the length of the column in air
The light waves transfer their oscillations and energy to other object in what way?
The frequency of the light waves physically causes the object to move
The energy of the light wave transfers to the electrons of the material which causes them to gain that energy
The wavelength of the light changes the velocity of the atoms in the material
The amplitude of the light will affect how much of it shines on the object
Answer: b the energy of light...
Explanation:
Your spaceship lands on a moon of a small planet that orbits a distant star. As you initially circled the moon, you measured its diameter to be 5480 km. After land you observe that a simple pendulum that had a frequency of 3.50 Hz on Earth now has a frequency of 1.82 Hz. What is the mass of the moon
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]
What is the velocity of an object that travels 200 meters in 20 seconds?
Answer:
10m/s
Explanation:
velocity=displacement/time
=200/20
=10m/s
In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested traveling at close to the speed of light. Suppose you wish to visit the red giant star Betelgeuse, which is 430 lyly away, and that you want your 20,000 kgkg rocket to move so fast that you age only 36 years during the round trip.
A. How fast (v) must the rocket travel relative to earth?
B. How much energy is needed to accelerate the rocket to this speed?
C. How many times larger is this energy than the total energy used by the United States in the year 2000, which was roughly 1.0 x 10^20 J?
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]
Current I flows along the positive z-direction in the inner conductor of a long coaxial cable and returns through the outer conductor. The inner conductor has radius a, and the inner and outer radii of the outer conductor are b and c, respectively. (a) Determine the magnetic field in each of the following regions: 0 ≤ r ≤ a, a ≤ r ≤ b, b ≤ r ≤ c, and r ≥ c. (b) Plot the magnitude of H as a function of r over the range from r = 0 to r = 10 cm, given that I = 10 A, a = 2 cm, b = 4 cm, and c = 5 c
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
A person is dragging a packing crate of mass 74.9 kg across a rough horizontal floor where the coefficient of kinetic friction is 0.35. He exerts a force F at and angle 43.0 degrees above the horizontal. What is the Force F such that the crate moves at a constant speed
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