[tex]F = 5.93×10^{13}\:\text{N}[/tex]
Explanation:
Given:
[tex]m_1= 2×10^{16}\:\text{kg}[/tex]
[tex]m_2= 4×10^{22}\:\text{kg}[/tex]
[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]
Using Newton's universal law of gravitation, we can write
[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]
[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]
[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt
This question is incomplete, the complete question is;
Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.
1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?
Fwith belt =
2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.
Fwithout belt =
Answer:
1) The Net force on the driver with seat belt is 10.3 KN
2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Explanation:
Given the data in the question;
from the equation of motion, v² = u² + 2as
we solve for a
a = (v² - u²)/2s ----- let this be equation 1
we know that, F = ma ------- let this be equation 2
so from equation 1 and 2
F = m( (v² - u²)/2s )
where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.
1)
Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.
i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m
so we substitute the given values into the equation;
F = 70( ((0)² - (18)²) / 2 × 1.1 )
F = 70 × ( -324 / 2.4 )
F = 70 × -147.2727
F = -10309.09 N
F = -10.3 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwith belt = 10.3 KN
Therefore, Net force of the driver is 10.3 KN
2)
No sit belt,
m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m
we substitute
F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )
F = 70 × ( -324 / 0.022 )
F = 70 × -14727.2727
F = -1030909.08 N
F = -1030.9 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwithout belt = 1030.9 KN
Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
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Galaxies that are 400 million light years away have a red shift of 0.03 approximately. A radio wave coming from one of these galaxies has an observed wavelength of 125 meters. What is the emitted wavelength in meters and the observed frequency in Megahertz
Answer:
The correct answer is "121.36 meters and 2.40 MHz".
Explanation:
Given:
Red shift,
[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]
Wavelength,
[tex]\lambda = 125 \ meters[/tex]
The observed frequency will be:
⇒ [tex]f = \frac{c}{\lambda}[/tex]
[tex]=\frac{3\times 10^8}{125}[/tex]
[tex]=24\times 10^5[/tex]
[tex]= 2.40\times 10^6[/tex]
[tex]=2.40 \ MHz[/tex]
hence,
The Emitted wavelength will be:
⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]
[tex]\frac{125}{\lambda_0}-1 =0.03[/tex]
[tex]125=1.03 \ \lambda_0[/tex]
[tex]\lambda_0=\frac{125}{1.03}[/tex]
[tex]=121.36 \ m[/tex]
Hi,A body changes its velocity from 60 km/hr to 72 km/hr in 2 sec.Find the acceleration and distance travelled.
Answer:
Initial velocity, u = 60 km/h = 16.7 m/s
Final velocity, v = 72 km/h = 20 m/s
time, t = 2 sec
From first equation of motion:
[tex]{ \bf{v = u + at}}[/tex]
Substitute the variables:
[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]
A 50.0 kg person is walking horizontally with constant acceleration of 0.25 m/s² inside an elevator. The elevator is also accelerating downward at a rate of 1.0 m/s². Sketch the path of the man as it is observed from someone on the ground. Explain your choice.
Answer:
The acceleration is in 2 D as in between east and south.
Explanation:
mass, m = 50 kg
acceleration, a = 0.25 m/s^2 horizontal
acceleration of elevator, a' = 1 m/s^2 downwards
When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:
An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x
Answer:
So the correct answer is letter e)
Explanation:
The electric field of an infinite yz-plane with a uniform surface charge density (σ) is given by:
[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]
Where ε₀ is the electric permitivity.
As we see, this electric field does not depend on distance, so the correct answer is letter e)
I hope it helps you!
In a robotics circuit, a voltage source of 75V is supplying a current, I to a series circuit of 5
resistances. Resistance, R1 = 5 KΩ and R2 = 10 KΩ. The voltage drops across 3 black boxes of
resistances R3 , R4 and R5 are 15V, 20V and 25V respectively. The current through the black
box of resistance, R5 is measured as 1mA. Calculate the voltage V1 and V2 across the
resistance R1 and R2 using the Voltage Divider Rule.
Answer:
In the given circuit, R
2
,R
6
and R
4
are in series. So,
R
1
′
=7+5+12=24Ω
Now R
1
′
and R
5
′
are in parallel. So,
R
2
′
1
=
8
1
+
24
1
=
24
3+1
=
24
4
=
6
1
R
2
′
=6ohm.
Now R
2
′
,R
1
and R
3
are in series. So,
R=R
2
′
+R
1
+R
3
=6+3+2=11ohm.
We know i=
R+r
E
=
11+1
6
=
12
6
=
2
1
i=0.5amp.
Consult Multiple-Concept Example 11 for background material relating to this problem. A small rubber wheel on the shaft of a bicycle generator presses against the bike tire and turns the coil of the generator at an angular speed that is 40 times as great as the angular speed of the tire itself. Each tire has a radius of 0.330 m. The coil consists of 163 turns, has an area of 3.23 x 10-3 m2, and rotates in a 0.0948-T magnetic field. The bicycle starts from rest and has an acceleration of 0.601 m/s2. What is the peak emf produced by the generator at the end of 6.72 s
Answer:
[tex]Emf=24.4V[/tex]
Explanation:
From the question we are told that:
Angular speed [tex]\omega=40*\omega'[/tex]
Radius [tex]r=0.33m[/tex]
No. Turns [tex]N=163[/tex]
Area [tex]A= 3.23 * 10^{-3} m^2[/tex]
Magnetic field. [tex]B=0.0948T[/tex]
Acceleration [tex]a=0.60m/s^2[/tex]
Time [tex]t=6.72s[/tex]
Generally the equation for momentum is mathematically given by
[tex]\omega'=\omega_o+\frac[a}{r}t[/tex]
[tex]\omega'=0+\frac{0.60}{0.33}*6.72[/tex]
[tex]W=12.22rad/s^2[/tex]
Therefore
[tex]\omega=Aw[/tex]
[tex]\omega=12.22*40[/tex]
[tex]\omega=488.7rads/s[/tex]
Generally the equation for Peak emf is mathematically given by
[tex]Emf=NBA \omega[/tex]
[tex]Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s[/tex]
[tex]Emf=24.4V[/tex]
True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.
Answer:
false?
Explanation:
The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.
Answer:
I think the answer is False.
3. If you change the resistance of the resistor:
a. How does the current through the circuit change? (answer, explain, evidence)
b. How does the voltage of the battery change? (answer, explain, evidence)
Answer:
Explanation:
Changing the resistance of a resistor means the resistance is either increased or decreased.
a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.
Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;
V = IR
6 = I x 30
I = 0.2 A
If the resistance is changed to 50 Ohm's, then:
I = 0.12 A
(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.
In the previous example, if the resistance is changed to 5 Ohm's, then:
V = IR
6 = I x 5
I = 1.2 A
(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.
An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10 m/s. (Assume the speed of sound is 343 m/s.)
(a) What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?
Hz
(b) What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?
Hz
Answer:
a) [tex]F=475.7Hz[/tex]
b) [tex]F'=410.899Hz[/tex]
Explanation:
From the question we are told that:
Velocity of eagle [tex]V_1=35m/s[/tex]
Frequency of eagle [tex]F_1=440Hz[/tex]
Velocity of Black bird [tex]V_2=10m/s[/tex]
Speed of sound [tex]s=343m/s[/tex]
a)
Generally the equation for Frequency is mathematically given by
[tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]
[tex]F=440(\frac{343-10}{343-35})[/tex]
[tex]F=475.7Hz[/tex]
b)
Generally the equation for Frequency is mathematically given by
[tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]
[tex]F'=440(\frac{343+10}{343+35})[/tex]
[tex]F'=410.899Hz[/tex]
What is total resistor formula
Answer:
If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.
Explanation:
explain why an equation may be homogenous with respect to its unit but still be incorrect
In the Homogenous equation with respect to its unit if the constants has been removed, then the equation is still homogenous but it is incorrect.
What is Homogenous equation?Homogenous equation is one of the type of the differential equations. These equation have zero on the right hand side of the equality sign. If the equation has the independent variable on the right side of equal sign then it is said to be non - homogenous equation. The homogenous and non - homogenous equations are the two types of linear equations.
An example of Homogenous equation is,
The formula for kinetic energy can be represented as,
K. E = 1/2 (mv²)
Here, 1/2 can be removed as it was a constant.
K. E = mv².
This equation can be homogenous but still it is incorrect.
Learn more about homogenous equation,
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Canopus, which is in the constellation of Carina and Argo Navis, is 310 ly away. You plan a sight-seeing vacation for Canopus and book a flight on the fastest spaceship in the universe, which travels at the speed of light. How many meters will you traverse to reach your vacation destination
Answer:
[tex]D'=2.933*10^{18}m[/tex]
Explanation:
From the question we are told that:
Distance [tex]D=310ly[/tex]
Where
[tex]1ly=9.46*10^{15}[/tex]
Generally the equation for meters will you traverse to reach your vacation destination is mathematically given by
[tex]D'=D*1ly[/tex]
[tex]D'=310*9.46*10^{15}[/tex]
[tex]D'=2.933*10^{18}m[/tex]
Which statement below is correct about this planet?
Everything on the planet will fall twice faster than on Earth.
Everything on the planet will fall twice slower than on Earth.
Everything on the planet will move twice faster than on Earth.
Everything on the planet will move twice slower than on Earth.
Answer:
Everything on the planet will move twice faster than on Earth.
Explanation:
Uranus has a mass of about 8.681 × 10^25 kg. That is its larger than earth and is also has a faster rotation peroid of 17 hours. Its has a gravitational potential less than earth.A child weighing 200 N is being held back in a swing by a horizontal force of 125 N, as shown in the image. What is the tension T in the rope that supports the swing in units of Newtons? Note: Please enter only the numerical answer. If you include any units in your answer, your answer will be counted as incorrect. T F= 125 N Weight = 200 N
Answer:
75
Explanation:
i am not sure but if 200N boy is being held back then the force that's holding him back must be equal to or greater than his weight. if 125N is already exerted then the tension will be:
T=200-125
= 75
A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?
Answer:
[tex]t=0.50cm[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lamda=3c[/tex]m
Refraction Index [tex]n=1.50[/tex]
Generally the equation for Destructive interference for Normal incidence is mathematically given by
[tex]2nt=m(\frac{1}{2})\lambda[/tex]
Since Minimum Thickness occurs at
At [tex]m=0[/tex]
Therefore
[tex]t=\frac{\lambda}{2}[/tex]
[tex]t=\frac{3}{4(1.50)}[/tex]
[tex]t=0.50cm[/tex]
how do I learn French fast for an examination
Explanation:
It isn't possible to learn an entire language fast. For an examination, memorise some of the easy words. Practice mew minutes speaking French and keep speaking it till you know it.
A pulse traveled the length of a stretched spring the pulse transferred...A)energy only B)mass only C)both energy and mass D) neither energy nor mass
Answer:
A
Explanation:
So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.
If you tethered a space station to the earth by a long cable, you could get to space in an elevator that rides up the cable much simpler and cheaper than riding to space on a rocket. There's one big problem, however: There is no way to create a cable that is long enough. The cable would need to reach 36,000 km upward, to the height where a satellite orbits at the same speed as the earth rotates; a cable this long made of ordinary materials couldn't even support its own weight. Consider a steel cable suspended from a point high above the earth. The stress in the cable is highest at the top; it must support the weight of cable below it.
What is the greatest length the cable could have without failing?
Answer:
[tex]l=12916.5m[/tex]
Explanation:
Distance [tex]d=3600km[/tex]
Since
Density of steel [tex]\rho=7900kg/m^3[/tex]
Stress of steel [tex]\mu= 1*10^9[/tex]
Generally the equation for Stress on Cable is mathematically given by
[tex]S=\frac{F}{A}[/tex]
[tex]S=\frac{\rho Alg}{A}[/tex]
Therefore
[tex]l=\frac{s}{\rhog}[/tex]
[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]
[tex]l=12916.5m[/tex]
Cho điện cực dưới điện cực trên . Hàm thế biến thiên theo qui luật:
Xác định sự phân bố điện tích khối và .
Answer:
Quantum theory gives the concept of
If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the track. Assume the track is frictionless. Hint: use the definitions of acceleration and Newton’s second law.
Answer:
hehe
Explanation:
I dont know because I am a noob ant study
The driver provides a constant force on the engine through the foot pedal. Eventually the van stops accelerating and reaches a constant speed.
c Explain why the van reaches a constant speed if the driver provides a constant driving force to the van.
It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is
∑ F = (force from engine) - (resistive forces) = 0
A hockey ball is flicked of the ground with initial velocity of 2.0m/s upwards and 10m/s horizontally. Calculate the distance travelled from the point where the ball is flicked and to the point where the ball hits the ground.
Answer:
imma try and fail again and again
A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{\rm total}=\Delta K[/tex]
or
[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]
Compute the work performed by friction:
[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]
[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]
[tex]\implies \boxed{\mu\approx0.45}[/tex]
The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
What is coefficient of friction?Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{total}=\Delta K[/tex]
or
[tex]W_{friction}+W_{spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]
Compute the work performed by friction:
[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{friction}=-f(0.20)[/tex]
[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]
[tex]\mu =0.45[/tex]
Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
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When an external magnetic flux through a conducting loop decreases in magnitude, a current is induced in the loop that creates its own magnetic flux through the loop. How does that induced magnetic flux affect the total magnetic flux through the loop
Answer:
Len's law
Explanation:
We can explain this exercise using Len's law
when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.
In which type of mixture do the physically distinct component parts each have distinct properties?
Answer:
In heterogeneous mixture do the physically distinct component parts each have distinct properties.
Explain how the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure. Sun Earth at position B Earth at position A
Answer:
The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.
Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloon and wall. Does the balloon move, if so which way
Answer:
Move towards the wall.
Explanation:
When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.
As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.
Consider an airplane with a total wing surface of 50 m^2. At a certain speed the difference in air pressure below and above the wings is 4.0 % of atmospheric pressure.
Required:
Find the lift on the airplane.
Answer:
[tex]F=202650N[/tex]
Explanation:
From the question we are told that:
Area [tex]a=50m^2[/tex]
Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]
Generally the equation for Force is mathematically given by
[tex]F=dP*A[/tex]
[tex]F=4053*50[/tex]
[tex]F=202650N[/tex]
