Two ships are docked next to each other. Their centers of mass are 39m apart. One ship’s mass is 9.2 *10^7 kg and the other ship’s mass is 1.84*10^8 kg. What gravitational force exists between them?
Please help!
Answer:
742.3N
Explanation:
Given parameters:
Distance = 39m
Mass 1 = 9.2 x 10⁷kg
Mass 2 = 1.84 x 10⁸kg
Unknown:
Gravitational force between the ships = ?
Solution:
To solve this problem, we apply the newton's law of universal gravitation:
Fg = [tex]\frac{G x mass 1 x mass 2}{r^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
r is the distance or separation
Fg = [tex]\frac{6.67 x 10^{-11} x 9.2 x 10^{7} x 1.84 x 10^{8} }{39^{2} }[/tex] = 742.3N
The chart below summarizes the forces applied to four different objects.
Which object will experience the greatest acceleration?
A. Z
B. X
C. Y
D. W
Answer:
C. Y
Explanation:
From Newton's second law of motion, we know that:
Force = mass x acceleration
So;
acceleration = [tex]\frac{Force }{mass}[/tex]
Therefore, to have the highest acceleration at a constant force, the mass must be low. Acceleration is inversely proportional to mass.
Y has the least mass and it will have the highest acceleration
Can a single atom be considered a molecule?
A:only if the atom is found in water
B:no, it takes two or more atoms bonded to create a molecule
C:only if it is an oxygen atom floating in the air
D:yes, all atoms are made up of many different molecules
In designing buildings to be erected in an area prone to earthquakes, what relationship should the designer try to achieve between the natural frequency of the building and the typical earthquake frequencies?
A) The natural frequency of the building should be exactly the same as typical earthquake frequencies.
B) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly lower
C) The natural frequency of the building should be very different frem typical earthquake frequencies
D) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly higher.
Answer:
C) The natural frequency of the building should be very different from typical earthquake frequencies
Explanation:
We shall apply the concept of resonance in this problem .
When a body is applied an external harmonic force ( forced vibration) such that natural frequency of body is equal to frequency of external force or periodicity of external force , the body vibrates under resonance ie its amplitude of vibration becomes very high .
In the present case if natural frequency of building becomes equal to the earthquake's frequency ( external force ) , the building will start vibrating with maximum amplitude , resulting into quick collapse of the whole building . So to avoid this situation , natural frequency of building should be very different from typical earthquake frequencies .
The second law of thermodynamics imposes what limit on the efficiency of a heat engine?
A. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy it extracts from a hot reservoir.
B. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy extracted as useful work.
C. A heat engine must deposit some energy in a cold reservoir.
Answer:
C. A heat engine must deposit some energy in a cold reservoir.
Explanation:
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."
This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.
Then we have the equation:
Q = W + q
From this we can conclude that the correct option is:
C. A heat engine must deposit some energy in a cold reservoir.
There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.
C. A heat engine must deposit some energy in a cold reservoir.
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir". This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir. Then we have the equation: Q = W + q There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.Therefore, option C is correct.
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The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr
Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Answer:
2 m/s²
Explanation:
From the given information:
The first mass m_1 = 0.6 kg
The second mass m_2 = 0.3 kg
The magnitude for the acceleration of 0.3 kg is:
a = net force/ effective mass
Mathematically, it can be computed as follows:
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]
[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]
a ≅ 2 m/s²
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.
Answer:
the tack's tangential speed is 5.59 m/s
Explanation:
Given that;
R = 0.331 m
wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so
ω = 2.69 rev/s × 2π/1s = 16.9 rad/s
using the relation of angular speed with tangential speed
tangential speed v of the tack is expressed as;
v = R × ω
so we substitute
v = 0.331 m × 16.9 rad/s
v = 5.59 m/s
Therefore, the tack's tangential speed is 5.59 m/s
A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
Two spherical objects have masses of 100 kg and 200 kg. Their centers are
separated by a distance of 40 cm. Find the gravitational attraction between
them.
Answer:
8.34 x 10⁻⁶N
Explanation:
Given parameters:
Mass 1 = 100kg
Mass 2 = 200kg
Distance of separation = 40cm = 0.4m
Unknown:
Gravitational force of attraction between them = ?
Solution:
To solve this problem, we use the expression below which is derived from the Newton's law of universal gravitation:
Fg = [tex]\frac{G x mass 1 x mass 2}{d^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
d is the separation
Now;
Fg = [tex]\frac{6.67 x 10^{-11} x 100 x 200}{0.4^{2} }[/tex] = 8.34 x 10⁻⁶N
In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 7.08 × 103 s and orbital speed of 3.40 × 103 m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 × 106 m. Calculate the mass of Mars.
Answer: [tex]5.944\times 10^{23}\ kg[/tex]
Explanation:
Given
Time period [tex]T=7.08\times 10^3\ s[/tex]
Orbital speed [tex]v=3.40\times 10^3\ m/s[/tex]
mass of GS [tex]m_{GS}=930\ kg[/tex]
Radius of Mars [tex]r=3.43\times 10^6\ m[/tex]
Consider the mass of mars is M
Here, Gravitational pull will provide the centripetal force
[tex]F_G=F_c[/tex]
[tex]\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}[/tex]
[tex]M=5.944\times 10^{23}\ kg[/tex]
In March 1999 the Mars Global Surveyor (GS) entered its final orbit on Mars, sending data back to Earth. The mass of Mars is approximately 6.419 × 10²³ kg.
Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit:
T² = (4π² / GM) × a³
In a circular orbit, the semi-major axis is equal to the radius of the orbit (r).
Given:
Orbital period (T) = 7.08 × 10³ s
Orbital speed (v) = 3.40 × 10³ m/s
Mass of GS (m) = 930 kg
Radius of Mars (r) = 3.43 × 10⁶ m
The orbital speed (v) is related to the radius (r) and the gravitational constant (G) by:
v = √(GM / r)
v² = GM / r
G = (v² × r) / M
T² = (4π² / [(v² × r) / M]) × r³
T² = (4π² × M × r²) / v²
M = (T² × v²) / (4π² × r²)
M = ( (7.08 × 10³)² × (3.40 × 10³)² ) / (4π² × (3.43 × 10⁶)²)
M = 6.419 × 10²³ kg
Therefore, the mass of Mars is approximately 6.419 × 10²³ kg.
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A pingpong ball has 2 kg/s of momentum when
thrown 8 m/s. Find the mass of the ball.
Answer:
0.25 kg
Explanation:
p = mv
2 = m(8)
2/8 = m(8)/8 *cancels
m = 1/4 OR 0.25 kg
A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
Let A be the second to last digit and let B be the last two digits of your 8-digit student ID. Example: for 20245347, A = 4 and B = 47.A ball rolls off a table. The table top is 1.2 m above the floor and the ball lands 3.6 m from the base of the table. Determine the speed of the ball at the time it rolled over the edge of the table? Calculate the answer in m/s and rounded to three significant figures.
Answer:
7.35 m/s
Explanation:
Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.
y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.
So, substituting the values of the variables into the equation, we have
y - y' = ut - 1/2gt²
0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²
- 1.2 m = 0 - (4.9 m/s²)t²
- 1.2 m = - (4.9 m/s²)t²
t² = - 1.2 m/- (4.9 m/s²)
t² = 0.245 s²
t = √(0.245 s²)
t = 0.49 s
Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.
So, d = vt
v = d/t
= 3.6 m/0.49 s
= 7.35 m/s
Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.
It is shown thus V = √(u² + v²)
= √(0² + v²)
= √(0 + v²)
= √v²
= v
= 7.35 m/s
A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the turnaround marker to the starting marker at 16 m/s. What is her average speed for the whole race?
Answer:
12.31 m/s
Explanation:
If we recall from the previous knowledge we had about speed,
we will know that:
speed = distance/ time.
As such:
The average speed of the rider bicycle is
average speed = total distance/ total time
Mathematically, it can be computed as:
[tex]v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}[/tex]
[tex]v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}[/tex]
[tex]\mathbf{v_{avg} =12.31 \ m/s}[/tex]
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which changes will increase the rate of reaction during combustion
Answer:
reducing temperature of the surrounding
Explanation:
combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer
If the force of gravity suddenly stopped acting on planets, they would
A.) spiral slowly towards the sun
B.) continue to orbit the sun
C.) move in straight lines tangent to thier orbits
D.) spiral slowly away from the sun
E.) fly straight away from the sun
I want you to think about each of these scenarios, what do you think will happen after? I just want you to think about it, and write a little about what is going to happen.
2. A truck is moving at 20 mph. Your car is standing still at a light and the truck crashes into you before the driver has a chance to step on the brakes.
3. You are driving your car at 20 mph. A bicycle right ahead of you suddenly stops and you crash into it before you have a chance to step on the brakes.
4. A bicycle is moving at 20 mph. Your car is stopped for a light and the bicycle crashes into you.
A ray of monochromatic light is incident on a plane mirror at and angle of 30. The angle of reflection for the light is
1)15
2)30
3)60
4)90
Answer:
30 degrees
Explanation: reflection, same angle
For a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
Reflection occurs when radiation bounces off from a surface. Light is an electromagnetic wave and it can be reflected. According to the laws of reflection, the angle of incidence is equal to the law of reflection.
Hence, for a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
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Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at the effective radius of the flywheel).
(a) How much work is done to bring this wheel from rest to a speed of 120 rev/min in a time. interval of 30.0 s?
(b) What is the applied torque on the fly-wheel (assumed constant)?
Answer:
A)5524J,
B) 29.2Nm
Explanation:
This question can be treated using work- energy theorem
Work= change in Kinectic energy
W= Δ KE
Work= difference between the final Kinectic energy and intial Kinectic energy.
We know that
Kinectic energy= 1/2 mv^2 .............eqn(1)
This can be written in term of angular velocity, as
KE= 1/2 I
PLS HELP ME!
A motorist is traveling 40ms-¹ and applies brakes and slow down at a rate of 2ms-² the available distance for the the motorist to stop is 400m will the motorist be able to stop?
Answer:
[tex] \underline{ \boxed{ yes}}\\[/tex]
Explanation:
[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]
To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
A charge of 7.1 x 10-4 C is placed at the origin of a Cartesian coordinate system. A second charge of 6.5 x 10-4 C lies 20 cm above the origin, and a third charge of 8.9 x 10-4 C lies 20 cm to the right of the origin. Determine the direction of the total force on the first charge at the origin. Express your answer as a positive angle in degrees measured counter clockwise from the positive x-axis.
Answer:
α = 36.21 °
β = 143.79°
Explanation:
To do this, we need to know the expression to calculate the angle.
In this case:
α₁ = tan⁻¹ (Fy₁/Fx₁) (1)
Now, let's analize the given data.
We have a charge q₁ at the origin of the cartesian coordinate system, so, it's at the 0. The charge q₂ is 20 cm above q₁, meaning is on the y-axis. Finally q₃ it's 20 cm to the right, meaning it's on the x-axis.
Knowing this,we can calculate the force that q₂ and q₃ are exerting over q₁. As these forces are in the x and y-axis respectively, we also are calculating the value of the forces in the x and y axis, that are needed to calculate the direction.
The expression to calculate the force would be Coulomb's law so:
F = K q₁q₂ / r² (2)
The value of K is 9x10⁹ N m² / C². Let's calculate the forces:
F₁₂ = Fy = 9x10⁹ * (7.1x10⁻⁴) * (6.5x10⁻⁴) / (0.020)²
Fy = 1.04x10⁷ N
F₁₃ = Fx = 9x10⁹ * (7.1x10⁻⁴) * (8.9x10⁻⁴) / (0.020)²
Fx = 1.42x10⁷ N
Now that we have both forces, we can calculate the magnitude of the force:
F = √(Fx)² + (Fy)²
F = √(1.04x10⁷)² + (1.42x10⁷)²
F = 1.76x10⁷ N
Finally, the direction would be applying (1):
α = tan⁻¹ (1.04x10⁷/1.42x10⁷)
α = 36.21 °
And counter clockwise it would be:
β = 180 - 36.21 = 143.79°
Hope this helps
WHat does that mean?
Many scientific studies have found that colds are caused by viruses. What is this? *
Fact
Interpretation
Analysis
Opinion
Answer:
Analysis
Explanation:
Because you must Analysis each and every cold too find out which virus caused this.
It’s weird because Interpretation and Analysis have the meaning of examination
A car is traveling at a constant speed of 20 m/s for 3 seconds. Then the driver puts on the brakes. The total distance the car travels is 100 m. What is the total time the car was moving?
Answer:
15 seconds
Explanation:
If car was moving at 20m/s for 3 sec.
if car traveled 100m = 15 sec total
a 1 mole of an ideal gas is kept at 0°C during expansion from 30l to 10l .How much work is done on the gas during expansion
Answer:
20 J
Explanation:
Work done is given force by distance .
W= F * d where F is force given by the product of pressure and area
W= P* Δv where Δv is change in volume.
Given that ;
1 mole of an ideal gas is kept at 0°C, the pressure of the gas is : 1 atm.
Δv is change in volume , 30 l - 10l = 20 l
W= 1 * 20 = 20 J
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.