Answer:
the required speed with which the missile move relative to the Earth is -0.727c
Explanation:
Given the data in the question;
relative velocity relation;
u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]
so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]
and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]
hence, u' is the speed of B as seen by A
so
u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]
now, given that; V[tex]_A[/tex] = 0.9c and V[tex]_B[/tex] = 0.5c
we substitute
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)
u' = ( -0.4c ) / 1 - 0.45
u' = -0.4c / 0.55
u' = -0.727c
Therefore, the required speed with which the missile move relative to the Earth is -0.727c
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
plz answer the question
Answer:
Ray A = Incidence ray
Ray B = Reflected ray
Explanation:
From the law of reflection,
Normal: This is the line that makes an angle of 90° with the reflecting surface.
Ray A is the incidence ray: This is the ray that srikes the surface of a reflecting surface. The angle formed between the normal and the incidence ray is called the incidence angle
Ray B is the reflected ray: This is the ray leaves the surface of a reflecting surface. The angle formed between the reflected ray and the normal is called reflected angle
Pete is investigating the solubility of salt (NaCl) in water. He begins to add 50 grams of salt to 100 grams of
room temperature tap water in a beaker. After adding all of the salt and stirring for several minutes, Pete
notices a solid substance in the bottom of the beaker. Which statement best explains why there is a solid
substance in the bottom of the beaker?
A. The salt he is using is not soluble in water.
B. The salt is changing into a new substance that is not soluble in water,
C. The dissolving salt is causing impurities in the water to precipitate to the bottom
D. The water is saturated and the remaining salt precipitates to the bottom
Answer:
would the answer be c
Explanation: that what i think in my opian
Answer:
A
Explanation:
Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous
Answer:
francium
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
help asap PLEASE I will give u max everything all that
steps if possible
Explanation:
2. [tex]R_T = R_1 + R_2 + R_3 = 625\:Ω + 330\:Ω + 1500\:Ω[/tex]
[tex]\:\:\:\:\:\:\:= 2455\:Ω = 2.455\:kΩ[/tex]
3. Resistors in series only need to be added together so
[tex]R_T = 8(140\:Ω) = 1120\:Ω = 1.12\:kΩ[/tex]
what is the frequency of a wave related to
Answer:
Frequency is the number of complete oscillations or cycles or revolutions made in one second.
A strong trough in a Rossby wave occurs when the jet stream A. bends towards the Equator. B. bends toward the poles. C. does not bend but maintains an east to west flow. D. does not bend but maintains a west to east flow.
Answer:
A. bends towards the Equator.
Explanation:
Rossby waves are inertial waves that are naturally occurring in a rotating fluids. These waves are also called as the planetary waves.
The Rossby waves are undulated that occur in the polar front jet stream when there is a significant differences in the temperatures between the polar and the tropical air masses.
It occurs when the polar air masses moves towards the equator and when the tropical air masses moves towards the pole. It is formed when the air bends away from the poles and bends towards the equator.
Hence the correct option is (A).
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?
Answer:
72 is the premimum of the insurance.
Explanation:
Below is the given values:
The loading = 0.4
Coinsurance rate = 0.2
Number of units = 100
Total number of units = 100 * 0.4 = 40
Remaining units = 60 * 0.2 = 12
Add the 60 and 12 values = 60 + 12 = 72
Thus, 72 is the premimum of the insurance.
A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initially at rest. Which of the following correctly identifies the change in total kinetic energy and the resulting speed of the objects after the collision? Kinetic Energy Speed
(A) Increases 2 m/s 3.2 m/s
(B) Increases Soold 2 m/s
(C) Decreases 3.2 m/s
(D) Decreases
Answer: (d)
Explanation:
Given
Mass of object [tex]m=2\ kg[/tex]
Speed of object [tex]u=5\ m/s[/tex]
Mass of object at rest [tex]M=3\ kg[/tex]
Suppose after collision, speed is v
conserving momentum
[tex]\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s[/tex]
Initial kinetic energy
[tex]k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J[/tex]
Final kinetic energy
[tex]k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J[/tex]
So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.
Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above
Answer:
Explanation:no change in surface tension
An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.
Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.
The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.
As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
Learn more here: https://brainly.com/question/15785205
A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.
Answer:
66.83 meters
Explanation:
After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.
1 mile = 1609.34 meters (multiply these meters by 65)
65 miles = 104,607 meters
1 hour = 3600 seconds
Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.
[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]
Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...
29.0575m * 2.3s = 66.83 meters
Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan
Answer:
F = 1010 Lb
the tension on the cable is greater than its resistance, which is why the plan is not viable
Explanation:
For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.
v = v₀ + a t
how the car comes out of rest v₀ = 0
a = v / t
let's reduce to the english system
v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s
let's calculate
a = 66/10
a = 6.6 ft / s²
now let's write Newton's second law
X axis
Fₓ = ma
with trigonometry
cos 20 = Fₓ / F
Fₓ = F cos 20
we substitute
F cos 20 = m a
F = m a / cos20
W = mg
F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]
let's calculate
F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20
F = 1010 Lb
Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.
