Answer:
the unit must always be indicated
Explanation:
displacement is alway in metre (m)
Answer:
A: measurement and direction
Explanation:
on edge! hope this helps!!~ (≡^∇^≡)
how does enormous energy get released from the sun
Answer:
By nuclear fission
Explanation:
The sun generates enormous energy through the process of nuclear fusion.
The core or the innermost part of the sun is characterized by high temperature and pressure. These two factors cause the separation of nuclei from electrons and the fusion of hydrogen nuclei to form a helium atom.
During the fusion process, energy is released.
Question 1 of 10
2 Points
What is the SI unit for gravitational potential energy?
A. Newton
B. Joule
C. Meter
D. Watt
Answer:
Joule
Explanation:
As any unit of energy (kinetic, potential, etc.), the gravitational potential energy has SI units of Joule. This s the work done by a force of 1 Newton to displace an object 1 meter in the direction of application of the force.
Describe the method you would use to measure the cicrumfrence of a cylinder using a thread and a meter
rule(4mk)
0)
Answer:
I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.
Explanation:
I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.
I then cut the thread at this length and extend it out on a table or floor.
I then place the length of the thread along the meter rule and mark of the position of its tip and bottom along the meter rule.
The difference between the bottom and top points is the circumference of the cylinder.
If the thread is longer than the meter rule, i mark off where the the length of the meter rule stops along the thread and then place the meter rule at that point and then mark off where the length of the thread stops.
The circumference of the cylinder is then 1 meter plus the extra measurement along the meter rule.
The magnitude of the electric potential depends on which factors?
O A. Mass and amount of charge
O B. Mass and distance
O C. Distance and amount of charge
O D. Density and amount of charge
Answer:
C
Explanation:
It is directly propotional to the amount of charge and inversly propotional to distance.
Answer:
C. Distance and amount of charge
Explanation:
The magnitude of the electric potential depends on which factors?
Distance and amount of charge
A locomotive moved 18.0 m [W] in a time of 6.00 s and stopped. After stopping, the locomotive moved 12.0 m [E] in 10.0 s. a. Determine the distance travelled by the locomotive. Show your work.
b. Determine the displacement of the locomotive. Show your work.
Answer:
a) The distance travelled by the the locomotive is 30 meters, b) The final displacement of the locomotive is 6 meters westwards.
Explanation:
a) The distance travelled is the sum of magnitudes of distances covered by the train during its motion. That is to say:
[tex]s_{T} = 18\,m + 12\,m[/tex]
[tex]s_{T} = 30\,m[/tex]
The distance travelled by the the locomotive is 30 meters.
b) The displacement is the vectorial distance of the train with respect to a point of reference, since west and east are antiparallel to each other, calculations can be simplified to a scalar form. Let suppose that movement to the east is positive. The calculations are presented below:
[tex]s_{P} = -18\,m +12\,m[/tex]
[tex]s_{P} = -6\,m[/tex]
The final displacement of the locomotive is 6 meters westwards.
Answer:
the dispalcement is 6m and distance travelled is 30m
Explanation:
helppppp me it's urgent please
the last one is the light will move in some manner not determined by the information here.
Light bends away from the normal, because it's moving from higher to lower refractive index.
Same bend-direction as when it goes from water into air.
Choose all correct statements: Refractive index of medium A with respect to medium B is 4/3 and refractive index of medium A with respect to medium C is 3/2. Medium B is denser than medium C When light travels from B to C light bends away from the normal Refractive index of C with respect to B is 8/9 Ratio of speed of light in medium B to the speed of light in medium A = 4/3
Answer:
Medium B is denser than medium C
When light travels from B to C light bends away from the normal .
Refractive index of C with respect to B is 8/9
Ratio of speed of light in medium B to the speed of light in medium A = 4/3
Explanation:
Refractive index of medium A / Refractive index of medium B = 4/3
Refractive index of medium A / Refractive index of medium C = 3/2
Dividing ,
Refractive index of medium B / Refractive index of medium C
= 3/2 x 3/4
= 9 / 8
Refractive index of medium B > Refractive index of medium C
Refractive index of C with respect to B is 8/9
speed of light in medium B / the speed of light in medium A
= refractive index of A / refractive index of B = 4 / 3
Un avión vuela a una velocidad de 900 km/h. Si tarda en viajar desde Canarias hasta la península 180 s ¿qué distancia recorre en ese tiempo?
Answer:
El avión recorrió 45 km en los 180 s.
Explanation:
La relación entre velocidad, distancia y tiempo se da de la siguiente manera;
[tex]Velocidad= \dfrac{Distancia}{Hora}[/tex]
Por lo cual los parámetros dados son los siguientes;
Velocidad = 900 km/h = 250 m / s
Tiempo = 180 s
Estamos obligados a calcular la distancia recorrida
De la ecuación para la velocidad dada arriba, tenemos;
Distancia recorrida = Velocidad pf viaje × Tiempo de viaje
Distancia recorrida = 900 km/h × 180 s = 900
Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km
Por lo tanto, el avión viajó 45 km en 180 s.
Study the image of earths layer which statement correctly compares the thicknesses of earths layers
This question is incomplete because the options are missing; here is the complete question:
Study the image of the Earth's layer which statement correctly compares the thicknesses of earths layers
A. Earth’s mantle is thinner than its oceanic crust.
B. Earth’s outer core is thicker than its mantle.
C. Earth’s continental crust is thicker than its lithosphere.
D. Earth’s lithosphere is thinner than its asthenosphere.
The answer to this question is D. Earth’s lithosphere is thinner than its asthenosphere.
Explanation:
The image shows the different layers that are part of Earth, as well as the thickness of each layer, in kilometers. In this, the thickest layer is the Mantle that is almost 2900 kilometers; this is followed in thickness by the outer and the inner core.
Additionally, other layers such as the continental/oceanic crust, the asthenosphere, and the lithosphere that are near the surface are thinner. About this, it can be concluded the lithosphere is thinner than the asthenosphere because the lithosphere has a thickness of 100 km, while the asthenosphere thickness is 660km. This makes option D the correct.
A bear can cover 60 min 4 seconds. What speed can it travel?
A. 25 m/s
B. 12 m/s
C. 15 m/s
D. 20 m/s
Answer:
C.15m/s
Explanation:
60 divided by 4 is 15
The mass and radious of the earth are 6*10^24 kg and 6.4*10^6 m respectively. calculate the acceleration due to gravity at the top of Mt. Everest of height 8,848 m from the sea level.
Answer:
9.74 m/s²
Explanation:
g = GM / r²
g = (6.67×10⁻¹¹ m³/kg/s²) (6×10²⁴ kg) / (6.4×10⁶ m + 8848 m)²
g = 9.74 m/s²
a Porsche 911 accelerates from rest to 27 metre per second due north in 5.8 seconds the mass of the car is 1400 kg what is the magnitude and direction of the average net force
Answer:
The average force has a magnitude 6524 N due north.
Explanation:
The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.
a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)
v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s
a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s
Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.
The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N
Since the acceleration is due north, the average force takes the direction of the acceleration.
So the direction of the average force is due north
The average force has a magnitude 6524 N due north.
If you travel at 60 miles per hour, how far will you travel in 2 hours?
A. not enough information
B. 45 miles
OC. 60 miles
D. 120 miles
Rachel and Andrew are riding on a ferry boat traveling east at 3.5 m/s. Rachel rolls a marble with a velocity of 0.95 m/s north, straight across the deck of the boat to Andrew. What is the velocity of the marble relative to the water?
Answer:
The velocity of the marble relative to the water is 3.63 m/s
Explanation:
The information given are;
The velocity of the ferry relative to the water = 3.5 m/s
Direction of motion of the ferry = East
The velocity of the marble (on the ferry) relative to the ferry = 0.95 m/s
The direction of the marble = North
The (resultant) velocity of the marble relative to the water, [tex]v_R[/tex], is found as follows;
Resolution of the velocities;
Velocity of the ferry = 3.5·i + 0·j
Velocity of the marble = 0·i + 0.95·j
[tex]v_R[/tex] = Combined (resultant) velocity = √(3.5² + 0.95²) = 3.63 m/s
The velocity of the marble relative to the water = 3.63 m/s.
please help
The statement "46 m, East" refers to the
of an object.
velocity
distance
displacement
speed
acceleration
Answer:
Displacement
Explanation:
Displacement is a vector quantity i.e it has both magnitude and direction. The direction is indicated with compass bearing i.e North, South, East or West or you could have North East, North West, South West, South East.
What is the kinetic energy of a 200kg boat moving at 2.7m/s?
Answer:
729 JoulesExplanation:
Kinetic energy is 1/2mv²
where m is the mass of the object
v is the velocity
From the question
m = 200kg
v = 2.7m/s
Kinetic energy is
1/2 × 200 × 2.7²
= 729 Joules
Hope this helps you
Using the formula W = mg, how many milliliters of water with a density of 1g/mL are required to weigh 0.75 newtons and g = 9.81 m/s2? Round to the nearest tenth. (Note: The mass will be in kg in your answer, thus convert to g and then to mL.) Answer: mL
Given that,
Weight = 0.75 N
Acceleration due to gravity = 9.81 m/s²
Density of water = 1 g/ml
We need to calculate the volume of water
Using formula of weight
[tex]W=mg[/tex]
[tex]W=\rho Vg[/tex]
[tex]mg=\rho\times V\times g[/tex]
Where, V = volume
g = acceleration
m = mass
Put the value into the formula
[tex]0.076\times 9.8\times10^3=1\times V\times9.8[/tex]
[tex]V= \dfrac{0.076\times 9.8\times10^3}{1\times9.8}[/tex]
[tex]V=76\ ml[/tex]
Hence, The volume of water is 76 mL.
Answer:
76.5mL
Explanation:
w = mg
0.75N = m * 9.81
m = 0.75 / 9.81 = 0.0765 kg
The mass in grams is 0.0765 * 1000 = 76.5g
At a density of 1 g/mL,
76.5g ÷ 1g/ml = 76.5mL
Two projectiles are thrown from the same point with the velocity of49ms-1. First is
projected making an angle with the horizontal and the second at an angle of (90- ). The
second is found to rise 22.5m higher than the first. Find the heights to which each will rise?
Answer:
Height of the first projectile = 49.98 m
Height of the second projectile = 72.52 m
Explanation:
From the given information;
Two projectiles are thrown from the same point with the velocity of49m/s
First is projected making an angle θ with the horizontal
and the second at an angle of 90 - θ.
Thus; for the first height to the horizontal; we have;
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex] ----- (1)
the second height in the vertical direction is :
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex] -----(2)
However; the second is found to rise 22.5 m higher than the first; so , we have :
[tex]\dfrac{v^2 cos^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
Let's recall that :
Cos²θ = 1 - Sin²θ
Replacing it into above equation; we have:
[tex]\dfrac{v^2}{2g} - \dfrac{v^2 sin^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]\dfrac{v^2}{2g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } \dfrac{v^2}{g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } - \dfrac {9.8 \times 22.5}{(49)^2} = sin^2 \theta[/tex]
[tex]\dfrac{1}{2 } - \dfrac {220.5}{2401} = sin^2 \theta[/tex]
[tex]sin^2 \theta= 0.408[/tex]
From (1);
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]H_1 = \dfrac{49^2 \times 0.408}{2*9.8}[/tex]
[tex]H_1 = \dfrac{979.608}{19.6}[/tex]
[tex]\mathbf{H_1 =49.98 \ m }[/tex]
Height of the first projectile = 49.98 m
Similarly;
From(2)
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex]
[tex]H_2 = \dfrac{v^2 (1-sin^2 \theta)}{2g}[/tex]
[tex]H_2 = \dfrac{49^2 (1-0.408 )}{2 \times 9.8}[/tex]
[tex]H_2 = \dfrac{2401 (0.592 )}{19.6}[/tex]
[tex]H_2 = \dfrac{1421.392}{19.6}[/tex]
[tex]\mathbf{H_2 = 72.52 \ m}[/tex]
Height of the second projectile = 72.52 m
NEED HELP!! WILL GIVE BRAINIEST!! When a driver presses the brake pedal, his car can stop with an acceleration of -5.4 meters per second squared. how far will the car travel while coming to a complete stop if its initial speed was 25 meters per second?
Answer:
d = 57.87 m
Explanation:
It is given that,
Acceleration of car due to applied brake is [tex]-5.4\ m/s^2[/tex]
Its initial speed was 25 m/s
Its final speed was 0 as its comes to rest.
Leat d is the distance covered by the car. It can be calculated using third equation of kinematics as :
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-5.4)}\\\\d=57.87\ m[/tex]
So, the distance covered by the car is 57.87 m.
Answer:
d = 57.87 m
Explanation: have a nice day :)
whats the texture of copper wire
Copper is a reddish orange, soft metal with a bright metallic luster. It is an excellent conductor of heat and electricity. Copper surfaces when exposed to air gradually turns to a dull, brownish color.
Los primeros vuelos de prueba del transbordador espacial usaban un "planeador" (con masa de 980 kg incluyendo el piloto). Después de un lanzamiento horizontal a 480 km/h a una altitud de 3500 m, el planeador finalmente aterrizaba con una rapidez de 210 km/h ¿Cuál era la fuerza promedio que la resistencia del aire ejercía sobre él cuando entraba con un ángulo de planeo constante de 12° con respecto a la Tierra?
Answer:
La fuerza promedio que ejerce la resistencia del aire sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra es 9613.8 N
Explanation:
Los parámetros dados son;
Masa del planeador = 980 kg.
Velocidad de lanzamiento horizontal = 480 km / h
Altitud de lanzamiento = 3500 m.
Velocidad de aterrizaje final = 210 km / h
Ángulo de deslizamiento, θ = 12 °
La ecuación para el equilibrio de fuerzas.
L · cos (θ) + D · sin (θ) = W
L · sin (θ) + D · cos (θ)
Dónde:
L = fuerza de elevación
D = fuerza de arrastre
W = Peso del "planeador"
W = Masa del "planeador" × Aceleración debido a la gravedad = 980 kg × 9.81 m / s²
W = 9613.8 N
Por lo tanto, tenemos;
L · cos (12) + D · sin (12) = 9613.8 ,,,,,,,,,,,,, (1)
L · sin (12) = D · cos (12) ,,,,,,,,,,,,, (2)
De la ecuación (2), tenemos;
L = D · (cos (12)/sin (12))
Sustituyendo el valor de L = D · (cos (12) / sin (12)) en la ecuación (1), se obtiene;
D · (cos (12)/sin (12)) · cos (12) + D · sin (12) = 9613.8
D = 9613.8/((cos (12)/sin (12)) · cos (12) + sin (12)) = 1998.82 N
L = D · (cos (12)/sin (12)) = 1998.82 · (cos (12)/sin (12)) = 9403.72 N
Por lo tanto, la fuerza da la fuerza promedio que la resistencia del aire ejerce sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra;
F = √ (L² + D²) = √ (9403.72² + 1998.82²) = 9613.8 N.
A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he enters the water, what average upward force did the water exert on him
Answer:
1120 N
Explanation:
The velocity with which he hits the water can be found with kinematics:
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)
v = -13.3 m/s
Or it can be found with conservation of energy.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × -9.8 m/s² × -9.00 m)
v = -13.3 m/s
Sum of forces on the diver after he hits the water:
∑F = ma
F − mg = m Δv/Δt
F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)
F = 1120 N
An airplane on a runway has an initial speed of 15 meters/second. What is its final velocity when it travels 600 meters on the runway with a constant acceleration of 6 meters/second2?
Answer:
86.17m/s
Explanation:
Since the motion is of constant acceleration, we can use one of the equations of motion as follows;
v² = u² + 2as
Where;
v = final velocity of the airplane
u = initial velocity of the airplane
s = distance covered by the plane
a = acceleration
From the question,
u = 15m/s
s = 600m
a = 6m/s²
Substitute these values into the equation above as follows;
v² = 15² + 2(6)(600)
v² = 225 + 7200
v² = 7425
v = √7425
v = 86.17m/s
Therefore, the final velocity of the airplane is 86.17m/s
If an object of 3.5 cm heitght is kept at a distance of 12 cm from the convex lens of focal length 8 cm. Find the position,nature and height of the height
Answer: The image is at a distance of 24 cm on the right of the lens, it will be an inverted, real image and has a height of 7 cm.
Explanation: To determine the position of the image of the object, use the Lens Equation, which is described by:
[tex]\frac{1}{f} = \frac{1}{u} +\frac{1}{v}[/tex]
where
f is focal length;
u is distance of the object from the lens;
v is distance of the image of the pbject from the lens;
Calculating the position of the image will be:
[tex]\frac{1}{v} =\frac{1}{f} - \frac{1}{u}[/tex]
[tex]\frac{1}{v} = \frac{1}{8} - \frac{1}{12}[/tex]
[tex]\frac{1}{v} = \frac{1}{24}[/tex]
v = 24 cm
Note that the distance of the image from the lens is positive. This means that the formed image is positioned at the right of the lens and is inverted.
For the height of the image, it can be used the Magnification (M) of an image:
M = [tex]\frac{h_{i}}{h_{o}} = \frac{v}{u}[/tex]
where:
[tex]h_{i}[/tex] is the height of the image
[tex]h_{o}[/tex] is the height of the object
For this image:
[tex]\frac{h_{i}}{3.5} = \frac{24}{12}[/tex]
[tex]h_{i} =[/tex] 2 * 3.5
[tex]h_{i}[/tex] = 7 cm
The image formed for this lens is 24 cm on the right of the lens, inverted related to the object and has height of 7 cm.
Which two substances could be produced in a combustion reaction?
A. Carbon dioxide and oxygen
B. Carbon dioxide and water
C. Heptane and water
D. Heptane and oxygen
Answer: B. Carbon dioxide and water
Explanation: Combustion simply refers to burning. A combustion reaction may thus be explained as a chemical process which involves the reaction between a substance and an oxidizer usually oxygen gas. The combustion reaction in most cases involves the reaction between hydrocarbons which are combustible compounds and oxygen gas. The reaction between the combustible compound and oxygen gas is then followed by the release of energy, with the product being carbondioxid and water. Below is an example of combustion reaction.
Combustion the Methane CH4.
Here, CH4 reacts with oxygen gas(O2) to produce carbondioxide and water.
CH4 + 2O2 ------> CO2 + 2H2O
Answer:
Correct answer B
Explanation:
Just took quiz
Which of the following is an example of the Doppler effect? A water bug on the surface of a pond is producing small ripples in the water. The bug is not moving. A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. Scientists measure the speed of light and find that it is constant. The gravitational pull of two objects on each other
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.
Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.
The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.
The phenomenon of Doppler effects is generally applicable to both sound and light.
An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.
Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc.
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
This should be the correct answer.
A ball is moving up an inclined plane, when it's velocity changes from 15m/s to 7m/s in 3 seconds . What is the acceleration.
Explanation:
[tex]a = \frac{change \: in \: velocity}{change \: in \: time} [/tex]
[tex]a = \frac{v - u}{t} [/tex]
[tex]v = 7 {ms}^{ - 1} [/tex]
[tex]u = 15 {ms}^{ - 1} [/tex]
[tex]t = 3 \: seconds[/tex]
[tex]a = \frac{7 - 15}{3} [/tex]
[tex]a = \frac{ - 8}{3} [/tex]
[tex]a = - 2.67 \: {ms}^{ - 2} [/tex]
21. A toy car starts from rest and accelerates at 1.50 m/s’ [E] for 5.25 s. What is the final
velocity, V, of the car? Show your work.
Answer: The final Velocity, V, of the car is 7.9m/s
Explanation:
Given the following :
Toy car starts from rest, therefore,
Initial Velocity (u) = 0
Acceleration (a) = 1.5m/s^2 E for time, t = 5.25s
Final velocity (V) of the car =?
Using the motion equation:
v = u + at
Where v = final Velocity
u = Initial Velocity
a = acceleration
t = time
v = 0 + 1.5(5.25)
v = 0 + 7.875
v = 7.875m/s
v = 7.9m/s
A roller coaster starts with a speed of 4.6 m/s at a point 49 m above the bottom of a dip.
Neglecting friction, what will be the speed of the roller coaster at the top of the next
slope, which is 29 m above the bottom of the dip?
Answer:
20.5m/s
Explanation:
neglecting friction the amount of energy at the bottom equals to the amount of energy at top. Law of conservation of energy.
ME= P.E+k.E
ME= mgh+ 1/2mv*v
ME = (m* 10 *49)+(1/2m(4.6*4.6))
MEtop=500.98m
MEbottom=MEtop: in order to obey the law of conservation of energy
MEbottom=mgh+ 1/2mv*v
500.98m=m*10*29+1/2m v* v
500.98m= m(290+1/2v*v) ...cancel m by m
500.98-290=1/2v*v criss cross multiple
210.98*2=v*v
v=√421.96
v=20.5 m/s
Example
Rain appears to fall vertically to a man walking at a rate of 3km/h. At
a speed of 6 km/h, it appears to meet him at an angle of 45° of vertical.
Find out the speed of rain.
Answer:
4.24 km/h
Explanation:
When the man walks at a rate of 3 km/h, the rain appears to fall vertically, which means it has the same horizontal rate, 3 km/h.
When the man walks at a rate of 6 km/h, the rain's horizontal velocity relative to the man is now 6 km/h − 3 km/h = 3 km/h. The rain meets the man at a 45 degree angle, so the vertical velocity is also 3 km/h.
Therefore, the total speed is:
v = √((3 km/h)² + (3 km/h)²)
v = 4.24 km/h