Answer:
ok
Explanation:
ok ok I will thanks for free
What famous equation describes the energy of stars?
Answer:
E=mc2
Explanation:
It explains that in their interiors, atoms fuse together, creating energy.
To solve this we must be knowing each and every concept related to energy and its different types. Therefore, the famous equation that describes the energy of stars is E=mc².
What is energy?In physics, energy is the capacity to accomplish work. It can be potential, kinetic, thermal, electric, chemical, radioactive, or in other forms.
There is also heat and work—energy inside the process of being transferred by one body to the other. Energy is always classified according to its type once it has been transmitted. As a result, heat transported could become thermal energy, whereas labor done may emerge as mechanical energy. The famous equation that describes the energy of stars is E=mc²
Therefore, the famous equation that describes the energy of stars is E=mc².
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Assume you are in the car and the car is moving at a certain speed to
school. Are you at rest or in motion with respect to the school? With
respect to the car?
Taking your hands off the wheel is an example of what type of driver distraction?
manual
visual
cognitive
adaptation
Answer:
manual
Explanation:
A manual distraction is when a driver takes their hand(s) off the wheel for any reason, for any amount of time. Without both hands on the wheel, your reaction time suffers, and so does your ability to steer.
А)
An automobile traveling along a straight road
increases its speed from 48 ft/s to 66 ft/s in
180 ft.
If the acceleration is constant, how much
time elapses while the auto moves the 180 ft?
Answer in units of s.
Answer:
4.5s
Explanation:
u=30m/s
v=50m/s
s=180m
a=constant(given)
By third equation of motion, v
2
=u
2
+2as
(50)
2
=(30)
2
+2a(180)
a=
3
40
m/s
2
By first equation of motion, v=u+at
50=30+(
9
40
)t
t=
2
9
=4.5sec
A piston has an external pressure of 15.0 atmatm. How much work has been done in joules if the cylinder goes from a volume of 0.120 liters to 0.450 liters
Answer: Work done is - 501.56 J
Explanation:
Given that;
External pressure P = 15.0 atm
Volume V1 = 0.120 liters
Volume V2 = 0.450 liters
Work done = ?
we know that; Work = -Pdv
where P is pressure and dv is change in volume
so we substitute our values into the equation
Work = -15.0 × ( 0.450 - 0.120)
= -15 × 0.33
= - 4.95 atm/L
we know that;
1 atm.L = 101.325 J
so
- 4.95 atm/L = 101.325 J × -4.95 atm/L ÷
= - 501.56 J
Therefore Work done is - 501.56 J
A rock climber is scaling rock wall at 10 m above the ground, and is moving downward at a speed of 0.25 m/s. If their phone falls out of their back pocket, how long before it shatters onto the ground?
Answer:
The time taken before the phone hits the ground is 1.4 s.
Explanation:
Given;
height of the rock wall, h = 10 m
speed of the climber, v = 0.25 m/s
The time taken before the phone hits the ground is given;
h = ut + ¹/₂gt²
10 = 0.25t + ¹/₂(9.8)t²
20 = 0.5t + 9.8t²
9.8t² + 0.5t - 20 = 0
solving this quadratic equation;
t = 1.4 s
Therefore, the time taken before the phone hits the ground is 1.4 s.
Ancient peoples both feared and revered Solar and Lunar Eclipses as they happened so rarely. Explain the following about eclipses:
1. Why are SOLAR eclipses more rare then lunar eclipses.
2. Explain what must occur for a SOLAR eclipse to occur.
3. Explain what must occur for a LUNAR eclipse to occur.
Answer:
Lunar and solar eclipses occur with about equal frequency. Lunar eclipses are more widely visible because Earth casts a much larger shadow on the Moon during a lunar eclipse than the Moon casts on Earth during a solar eclipse. As a result, you are more likely to see a lunar eclipse than a solar eclipse.
For a total eclipse to take place, the sun, moon and Earth must be in a direct line. The second type of solar eclipse is a partial solar eclipse. This happens when the sun, moon and Earth are not exactly lined up. The sun appears to have a dark shadow on only a small part of its surface.
For a lunar eclipse to occur, the Sun, Earth, and Moon must be roughly aligned in a line
Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 Newtons on earth, which has more mass?
a. Object a
b. Object b
c. Same mass
d. Other
Answer:
a. Object A
Explanation:
The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.
The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.
So that,
For object A on the moon, mass = 25 lbs
For object B on the earth, W = 25 N,
W = m x g
25 = m x 10 (g = 10 m/[tex]s^{2}[/tex])
m = [tex]\frac{25}{10}[/tex]
= 2.5 lbs
Mass of object B is 2.5 lbs.
Therefore, the mass of the object A is more than that of B.
Terell lives in Washington, D.C., where he can see the Potomac River from his bedroom window. When he sees this river, Terell thinks of swimming, boating, fun, and friends. What type of mental image is this?
A. analogical
B. symbolic
C. both analogical and symbolic
D. representative
Answer:
B. symbolic
Explanation:
The mental image formed in Terell's mind is a symbolic mental image.
A symbolic mental image helps to bring perception to imagery formed on the mind.
This process is symbolic because Terell can see that river ahead. If he cannot see the river then it is more of an analogical fantasy. The thoughts are predicated on the image of the River seen. Therefore, it is right to conclude that this is a symbolic mental image.Answer:
Terell lives in Washington, D.C., where he can see the Potomac River from his bedroom window. When he sees this river, Terell thinks of swimming, boating, fun, and friends. What type of mental image is this?
symbolic
Explanation:
A ball has a diameter of 3.77 cm and average density of 0.0839 g/cm3. What force is required to hold it completely submerged under water?
magnitude _________ N
The force required to hold it completely submerged under water is 0.252 N
As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.
Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation
F = Buoyant force - weight of sphere
Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m
Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³
Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg
Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N
Volume of water displaced = 4/3 π r³ = 2.805 e-5
Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N
F = 0.275 - 0.023 = 0.252 N
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The force required to hold it completely submerged under water is 0.25 N
The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³
The density of water [tex]\rho_w[/tex] = 1000 kg/m³
Diameter = 3.77 cm = 0.0377 m
radius of ball = 0.0377/2 = 0.01885 m
The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]
Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:
The force is required to hold it completely submerged under water (F) is:
[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]
F = 0.25 N
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An amateur golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00360 s. After the collision, the ball leaves the club at a speed of 34.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?
Answer:
The average force ≅ 519.44 N.
Explanation:
Impulse = change in momentum of a body
i.e Ft = m(v - u)
where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.
m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s
So that,
F x 0.00360 = 0.055(34 - 0)
F x 0.00360 = 0.055 x 34
= 1.87
F = [tex]\frac{1.87}{0.0036}[/tex]
= 519.4444
The average force exerted on the ball by the club is approximately 519.44 N.
A 600 kg rocket sled can be accelerated at a constant rate from rest to 1400 km/h in 2.1 s. What is the magnitude of the required net force
Answer:
7.2 × 10^5 N
Explanation:
The first step is to convert 1400 km/hr to m/s
= 1,400×1000/3600
= 1,400,000/3600
= 388.88 m/s
The acceleration can be calculated as follows
a= v-u/t
= 388.88/2.1
= 185.18
Therefore the required net force can be calculated as follow
= 388.88 × 185.18
= 7.2 × 10^5 N
what are the 3 main states of matter in water?/
Answer:
ice (solid), water (liquid) and vapor (gas)
Explanation:
How do you calculate the net force when there are multiple forces in different directions?
To find [tex]F_{net}[/tex] we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force [tex]F_{net} = \sqrt{(3)^2+(4)^2} \\[/tex] which gives [tex]F_{net} = 5[/tex]. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using [tex]\theta = \tan^{-1} (\frac{4}{3})[/tex] where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.
Please help me with this question guys.
Answer:
The average speed is 22.2 km/h
Explanation:
Average Speed
Given an object travels a total distance d and took a total time t, then the average speed is:
[tex]\displaystyle \bar v=\frac{d}{t}[/tex]
The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:
[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]
Then he drives d2=7 km at v2=43 km/h taking a time of:
[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]
The total time is
t=0.467 h + 0.163 h = 0.63 h
The total distance is
d = 7 km + 7 km = 14 km
The average speed is:
[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]
The average speed is 22.2 km/h
A 5-newton force directed east and a 5-newton force directed
north act concurrently on a point. Calculate the resultant of the two forces.
Answer:
[tex]5\sqrt{2} \ N[/tex]
Explanation:
It is given that,
A 5-newton force directed east and a 5-newton force directed north.
We need to find the resultant of the two forces.
As the two forces are acting in perpendicular to each other. The resultant of two such forces is given by :
[tex]F=\sqrt{5^2+5^2} \\\\=5\sqrt{2} \ N[/tex]
So, the resulatnt force is [tex]5\sqrt{2} \ N[/tex]
who plays adopt me on rob.lox!! plz answer people only if you play though tell me your best pet and your user name so we can play!!
Answer: I don't do adopt me but I do have a Ro.blox username its Nvm i cant but my user name
Explanation:
Answer:
adopt me sucks but my best pet was an albino monkey I think or a king monkey
Explanation:
uses of atmospheric pressure
What is the deBroglie wavelength of an electron moving at 1.37 x 106 m/s if the mass of the electron is 9.11 x 10-28 g
Answer:
The value is [tex]\lambda = 5.30 *10^{-10 } \ m[/tex]
Explanation:
From the question we are told that
The velocity of the electron is [tex]v = 1.37 *10^{6} \ m/s[/tex]
The mass of the electron is [tex]m = 9.11 *10^{-28} \ g = 9.11 *10^{-31} \ kg[/tex]
Generally the deBroglie wavelength is mathematically represented as
[tex]\lambda = \frac{h}{mv}[/tex]
Here h is the Planck'c constant with value [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
So
[tex]\lambda = \frac{6.62607015 * 10^{-34} }{9.11 *10^{-31} * 1.37 *10^{6}}[/tex]
=> [tex]\lambda = 5.30 *10^{-10 } \ m[/tex]
This is my question
Answer:
baby yoda
Explanation:
A heavy storm leaves 1 cm of rain on a city 5 km wide and 8 km long on a 2 hour period. How many cm3 of water fell on the city?
Answer: wow that’s hard
Explanation:
The above photo shows Plates A and B at an mid ocean ridge. If you were to draw arrows on the plates showing plate movement, which way would you draw arrows?
A . The arrows on plates A and B would show the plates moving in the same direction
B . The arrows on plates A and B would show the plates moving towards eachother
C . The arrows on plates A and B would show the plates moving in random directions
D . The arrows on plates A and B would show the plated moving away from each other
Answer:
D . The arrows on plates A and B would show the plated moving away from each other
Explanation:
The correct arrow in this instance will show that the plates A and B would be moving away from each other.
At the mid-ocean ridge, two plates are moving apart and pulling away from one another.
In short, at the mid-oceanic ridge, we have a divergent margin. At a divergent margin, the plates are spreading or moving apart.A force F = (2xî + 4yĵ), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x-direction from the origin to x = 4.99 m. Find the work W = ∫ F · dr done by the force on the object (in J).
Answer:
Work done, W = 24.9001 J
Explanation:
Given that,
Force, F = (2xî + 4yĵ) Where x and y are in meters
This force acts on an object as the object moves in the x-direction from the origin to x = 4.99 m.
We need to find the work done by the force on the object. It is given by :
[tex]W=\int\limits^a_b {F dr} \\\\W=\int\limits^{4.99}_0 {(2xi+4yj)dx} \\\\W=\int\limits^{4.99}_0 {2xi\ dx} \\\\W=x^2|_0^{4.99}\\\\\text{Applying limits}\\\\W=(4.99)^2-0^2\\\\W=24.9001\ J[/tex]
So, the work done is 24.9001 J.
1.) There was an earthquake in Salt Lake City, Utah, on March 18, 2020, in the morning at 9 hours, 9 minutes, and 45 seconds Mountain Standard Time (9:9:45 MST). If the velocity of the p-wave is 7.3 km/sec and the velocity of the s-wave is 5.1 km/sec and the s-p time lag is 16 seconds, what is the distance in kilometers from Salt Lake City to the focus of the earthquake? Explain how you calculated the answer.
Answer:
7 because salt lake and Southis weat
The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point located 0.40 L from the end
Answer:
The value is [tex]I = 0.0932 ML ^2[/tex]
Explanation:
From the question we are told that
The rotational inertia about one end is [tex]I_R = \frac{1}{3} ML^2[/tex]
The location of the axis of rotation considered is [tex]d = 0.4 L[/tex]
Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is [tex]0.4 M[/tex]
Generally the length of the rod from the its beginning to the axis of rotation consider is
[tex]k = 1 - 0.4 L = 0.6L[/tex]
Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is
[tex]m = 1- 0.4 M = 0.6 M[/tex]
Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is
[tex]I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
[tex]I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2[/tex]
Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is
[tex]I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is
[tex]I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
=> [tex]I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ][/tex]
=> [tex]I = 0.0932 ML ^2[/tex]
which statement is true of the particles that make up a substance? a. particles in a liquid have more energy than particles in a gas. b. particles in a liquid and particles in a solid have the same amount of energy. c. particles in a gas have more energy than particles in liquid. d. particles in a solid have moe energy than particles in a gas.
Explanation:
so sorry
don't know but please mark me as brainliest please
A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
Answer:
Explanation:
To get the focal length, we will use the lens formula;
1/f = 1/u + 1/v
f is the focal length
u is the object distance
v is the image distance
Given
since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.
u = 35cm
v = -2.0
1/f = 1/35-1/2
1/f = 2-35/70
1/f = -33/70
f = -70/33
f = -2.12 cm
f = -0.0212m
Power of a lend is the reciprocal of its focal length
Power of the lens = 1/f
P = 1/-0.0212
P = -47.17dioptres
The power of the lens is -47.17D
At what speed does a 1500 kg compact car have the same kinetic energy as a 18000 kgkg truck going 21 km/hr km/hr
Answer:
speed = 72.75km/hr
Explanation:
Let us first determine the Kinetic energy (KE) of the 18000kg truck
[tex]KE = \frac{1}{2} mv^2\\where:\\m = mass = 18000\\v = velocity = 21km/hr\\\therefore KE = \frac{1}{2} \times 18000 \times (21)^2\\= 9000 \times 441\\KE = 3,969,000\ Joules\\= 3,969\ KJ[/tex]
Next we will substitute this value of kinetic energy into the KE equation for the 1500kg compact car
[tex]KE = \frac{1}{2} mv^2\\3969000 = \frac{1}{2} \times 1500 \times v^2\\3969000 = 750 \times v^2\\v^2 = \frac{3969000}{750} = 5292\\v = \sqrt{5292} \\v = 72.75km/hr[/tex]
The speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h
The kinetic energy of an object is known to be the energy at work and in motion. It is usually expressed as;
[tex]\mathbf{K.E = \dfrac{1}{2}mv^2}[/tex]
To determine the speed at which the car will have the same kinetic energy as the truck, we can say that:
[tex]\mathbf{\implies \dfrac{1}{2}m_c v_c^2= \dfrac{1}{2}m_t v_t^2}[/tex]
[tex]\mathbf{m_c}[/tex] = mass of the car[tex]\mathbf{v_c}[/tex] = speed of the car[tex]\mathbf{m_t}[/tex] = mass of the truck[tex]\mathbf{v_t}[/tex] = speed of the truck∴
[tex]\mathbf{\implies \dfrac{1}{2}\times 1500\times v_c^2= \dfrac{1}{2}\times 18000 \times 21^2}[/tex]
[tex]\mathbf{v_c^2= \dfrac{3969000}{ 750}}[/tex]
[tex]\mathbf{v_c^2=5292}[/tex]
[tex]\mathbf{v_c=\sqrt{5292}}[/tex]
[tex]\mathbf{v_c\simeq 73 km/h}[/tex]
Therefore, we can conclude that the speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h
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C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
Answer:
The mass of the cart is 150 kg.
Explanation:
Given that,
Mass of a boy, m₁ = 50 kg
Initial speed of boy, u₁ = 10 m/s
Initial speed of car, u₂ = 0 (at rest)
The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s
Let m₂ is the mass of the cart. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]
So, the mass of the cart is 150 kg.
Running with an initial velocity of 13 m/s , a horse has an average acceleration of -1.88 m/s2 . How long does it take for the horse to decrease its velocity to 6.3 m/s
pls :/
Answer:
3.56s
Explanation:
Ya that's the right answer.
