Answer: 0.25m/s
Explanation:
Answer:
Explanation:
Blab balh its 0.25 m/s
A water skier is towing by motorboat at a constant velocity of magnitude 15 km /h. The boat speed up, and after a short interval the skier is towed at a new constant velocity of magnitude 20 km/h. What is the net force on the skier when she is moving at 15 km/h? And at 20 km/h?
Answer:
The skier is experimenting a net force of 0 newtons in both cases. ([tex]v_{1} = 15\,\frac{km}{h}[/tex], [tex]v_{2} = 20\,\frac{km}{h}[/tex])
Explanation:
According to Newton's First Law, an object is in equilibrium when it is either at rest or moving at constant velocity, which means that net force is equal to 0 newtons.
Therefore, the skier is experimenting a net force of 0 newtons in both cases.
A force of 75 N at an angle of 15° to the direction of motion moves a chair 3 m. Which change would result in more work being done on the chair?
Answer:
Decreasing the angle to 10
Explanation:
Edge 2020
At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2. What can the staff member estimate for the original speed of the race car if it came to a stop during the skid?
Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = [tex]\sqrt{734.22}[/tex]
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
Protons are found only in the atomic nucleus.
TRUE
FALSE
Answer:
true
Explanation:
Protons and neutrons are found in the nucleus
List out Units For: Acceleration
Unit for Acceleration is m/s² or ms raised to -2
A dog is 60 m away while moving at a constant velocity of 10 m/s towards you. How Long will it be before the dog is close enough to your face?
Answer:
6 seconds
Explanation:
The answer is 6 because if the dog is going 10 m/s it will take 6 seconds to get from 60 meters away to you.
1 point
1)A 1 kg brick is carried to up the Eiffel Tower to the 275 m platform. How
much gravitational potential energy does it acquire?
Answer:
The gravitational potential energy is 2,750 Joules
Explanation:
Given the following data;
Mass of brick = 1kg
Height of Eiffel Tower = 275m
Potential energy can be defined as the energy possessed by an object due to its position.
Mathematically, potential energy is;
[tex] P = mgh[/tex]
Also, we know acceleration due to gravity = 10m/s2
[tex] P = mgh[/tex]
[tex] P = 1 * 275 * 10 [/tex]
P = 2750 Joules.
Hence, the gravitational potential energy the brick acquired is 2750 Joules.
what does the galaxy made of ?
If you run east for 38m, stop suddenly, run another 27m east, then you run west for 47
m, what is your total displacement?
how to answer: #m, direction
Answer:
Total Displacement = -18m
Explanation:
Given
East Movement = 38m and 27m
West Movement = 47m
Required
Determine the total displacement
To answer this, we consider east as negative direction and the west as positive (using logics of number line).
Total Displacement = East Movement + West Movement
Total Displacement = -38m - 27m + 47m
Total Displacement = -18m
Hence, the total displacement is 18m towards East
how much force is needed to accelerate 300 kg at a rate of 4 m/s/s
Answer:
1200
Explanation:
Force = Mass x Acceleration
Answer:
[tex]\boxed {\tt 1,200 \ Newtons }[/tex]
Explanation:
Force can be found by multiplying the mass by the acceleration.
[tex]F=m*a[/tex]
The mass is 300 kilograms.
The acceleration is 4 meters per second squared.
[tex]m= 300 \ kg\\a= 4 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]F= 300 \ kg * 4 \ m/s^2[/tex]
Multiply.
[tex]F= 1200 \ kg * m/s^2[/tex]
1 Newton is equal to 1 kilograms meters per second squared, so our current answer of 1200 kg m/s² is equal to 1200 Newtons.
[tex]F= 1200 \ N[/tex]
The force needed is 1,200 Newtons.
Which model of the atom has electrons traveling in specific paths around the nucleus? Bohr's model Rutherford's model Thomson's model Dalton's model
Answer:
A, Bohr's model
Explanation:
Took the quiz, hope it's right for you.
(The other three did have something to do with it they just didn't create that specific model)
The model of the atom that has electrons traveling in specific paths around the nucleus is Rutherford's model.
Therefore, the correct answer is option B
J. J Thomson proposed that the atom is a sphere of positively charged matter in which negatively charged electron are embedded.
In Rutherford's model of atomic structure, the electrons moves in orbits by electrostatic attraction to the positively charged nucleus.
According to Dalton's atomic theory, atom were indestructible and indivisible solid particle.
Bohr's model made assumption that electron can only exist in circular orbits of definite quantum energy.
According to the question, the model of the atom that has electrons traveling in specific paths around the nucleus is Rutherford's model.
Therefore, the correct answer is option B
Learn more here : https://brainly.com/question/16776207
How are velocity, wavelength, and frequency related
Answer:
The wave velocity and the wavelength are related to the wave's frequency and period by vw=λT or vw=fλ. The time for one complete wave cycle is the period T. The number of waves per unit time is the frequency ƒ. The wave frequency and the period are inversely related to one another.
Explanation:
please help im dying
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It starts at the 14-inch mark, crawls to the 20-inch mark, then moves to the 16-inch mark. What was the total distance the ant traveled?
Given :
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west.
It starts at the 14-inch mark, crawls to the 20-inch mark, then moves to the 16-inch mark.
To Find :
The total distance the ant traveled.
Solution :
Total distance travelled by ant = (distance between 14 and 20 inch mark) +
(distance between 20 and 16 inch mark)
Total distance = (20-14 ) + ( 20-16) = 6 + 4 = 10 inch.
Therefore, total distance the ant traveled is 10 inch.
Hence, this is the required solution.
5 grams is thes than greater than or equal too 508mg
Answer:
yes 5 is greater
Explanation:
Answer: 5 grams is greater than 508 milligrams. Even though the number '508' is larger than '5,' you need to look at the units.
The moon euphoria is in a circular orbit around Jupiter. Which of the following describes the direction of the centripetal force acting on euphoria.
A) pointing hand into the path of its orbit
B) pointing towards the sun
C)Pointing away from Jupiter
D) pointing towards Jupiter
What will be acceleration due to gravity of a feather and coin if
they are dropped in vacuum?
Explanation:
Hey there!
We know that the objects in vacuum falls equally on the surface or bottom. As there is no air resistance in vacuum, the objects has equal stabdard acceleration. (i.e g = 9.8m/s^2)
So, the acceleration due to gravity is 9.8m/s^2.
Hope it helps....
Dos pilotos suicidas, que están inicialmente a una distancia de 500 m entre sí, deciden chocar directamente de frente arrancando ambos desde el reposo. Ambos autos pueden desarrollar una aceleración máxima constante de 15 m/s2 . Si el piloto A arranca un segundo antes que el piloto B, encuentra: a) la posición donde los autos chocan, medida a partir de la posición donde arranca el piloto A, y b) la rapidez relativa de la colisión (la rapidez de B con respecto a A justo antes de la colisión, ó viceversa).
Answer:
a) El punto de colisión de los dos automóviles desde donde parte el conductor del automóvil A es de aproximadamente 293,14 metros.
b) La velocidad del automóvil A, en relación con la velocidad del automóvil B, es de 15 m / s
Explanation:
Los parámetros del movimiento son;
La distancia entre ambos coches = 500 m
La aceleración de ambos coches = 15 m / s²
La dirección de movimiento de ambos coches = uno hacia el otro
La hora de inicio del conductor A = Un segundo antes de la hora de inicio del conductor B
Por lo tanto, de la ecuación de movimiento, tenemos;
s = u · t + 1/2 · a · t²
v² = u² + 2 · a · s
Dónde;
u = La velocidad inicial de los autos = 0 (los autos parten del reposo)
t = El tiempo de movimiento de una aceleración dada
a = La aceleración = 15 m / s²
s = La distancia recorrida en el tiempo t
Por lo tanto, para el controlador A, tenemos;
s₁ = 0 × (t + 1) + 1/2 × 15 × (t + 1) ² = 7.5 × (t + 1) ²
s₁ = 7.5 × (t + 1) ²
Para el conductor B, tenemos;
s₂ = 0 × t + 1/2 × 15 × t² = 7.5 × t²
s₂ = 7,5 × t²
Dado que ambos chocan a lo largo del camino de 500 m, tenemos;
s₁ + s₂ = 500 metros
∴ 7.5 × (t + 1) ² + 7.5 × t² = 500
∵ s₁ + s₂ = 7.5 × (t + 1) ² + 7.5 × t²
Lo que da;
15 · t² + 15 · t + 7.5 = 500
15 · t² + 15 · t - 492,5 = 0
Resolver usando la aplicación en línea da; t = -6,25181 o t = 5,25181
Dado que t es un número natural, tenemos el valor correcto para t = 5.25181 segundos
a) Por tanto, el punto de colisión de los dos coches desde donde parte el conductor del coche A es;
s₁ = 7.5 × (t + 1) ² = 7.5 × (5.25181 + 1) ² ≈ 293.14 metros
El punto de colisión de los dos automóviles desde donde parte el conductor del automóvil A ≈ 293,14 metros
b) La semilla de A en el punto de colisión se da de la siguiente manera
Velocidad, v₁ = u + a × (t + 1) = 0 + 15 × (5.25181 + 1) ≈ 93.78
v₁ ≈ 93,78 m / s
La semilla de B en el punto de colisión también se da de la siguiente manera
Velocidad, v = u + a × (t) = 0 + 15 × 5.25181 ≈ 78.78
v₁ ≈ 78,78 m / s
Por lo tanto, la velocidad del automóvil A, en relación con la velocidad del automóvil B, [tex]v_{relative}[/tex] se da como sigue;
= v₁ - v₂ = 93,78 m / s - 78,78 m / s = 15 m / s
la velocidad del automóvil A, relativa a la velocidad del automóvil B = 15 m/s.
The photo shows dew drops that formed on the grass when temperatures
cooled during the night. Which change of state happened to form the dew?
A. Liquid to gas
B. Gas to liquid
C. Solid to gas
D. Gas to solid
How do electromagnetic waves affect you?
Answer:
At low frequencies, external electric and magnetic fields induce small circulating currents within the body. ... The main effect of radiofrequency electromagnetic fields is heating of body tissues. There is no doubt that short-term exposure to very high levels of electromagnetic fields can be harmful to health.
Question 4 of 20
Luke wraps a magnetized steel nail with several coils of insulated wire and
then connects the loose ends of the wire to a 10 W lightbulb. Which step
could he take next to get the lightbulb to light up?
O A. Replace the magnetized nail with a nonmagnetic metal to cause
an electric current to flow in the wire.
O B. Remove the magnetized nail from the wire coil to induce a
magnetic field that will flow through the wire.
C. Move the magnetized nail back and forth within the wire coil to
induce an electric current in the wire.
D. Add several more coils of the wire around the magnetized nail to
make a stronger electric current.
SUBMIT
Answer:
C. Move the magnetized nail back and forth within the wire coil to induce an electric current in the wire.
Hope this helps.
#7 If the Moon’s mass was suddenly tripled, what would occur to the gravitational force between the Earth and the Moon?
(A) Gravitational force would increase by a factor of 3.
(B) Gravitational force would decrease by a factor of 3.
(C) Gravitational force would increase by a factor of 9.
(D) Gravitational force would decrease by a factor of 9.
Answer:
well it would be (A) Gravitational force would increase by a factor of 3.
Explanation:
because the moon is the only thing that gained mass not the Earth
A toy car rolls down a hill starting from rest. After 4.2 seconds its velocity is 18 meters per second. What is the acceleration of the car ?
SOMEONE PLZ HELP :(
WILL MARK BRAILIEST
Answer: if this is the same question I think it’s 24
Explanation:
What 2 gases make up most of the atmosphere?
Answer:
nitrogen oxygen carbon dioxideand argon
Explanation:
A student fires a cannonball vertically upwards. The cannonball returns to the
ground after a 4.60s flight. Determine all unknowns and answer the following
questions. Neglect drag and the initial height and horizontal motion of the
cannonball. Use regular metric units (ie. meters).
How long did the cannonball rise?
unit
What was the cannonball's initial speed?
unit
What was the cannonball's maximum height?
unit
V
Answer:
(a). The distance is 207 m.
(b). The initial velocity is 45.0 m/s
(c). The maximum height is 103.3 m
Explanation:
Given that,
Time = 4.60 s
We need to calculate the initial velocity
Using equation of motion
[tex]v=u-gt[/tex]
Put the value into the formula
[tex]0=u-9.8\times4.60[/tex]
[tex]u=45.0\ m/s[/tex]
We need to calculate the distance
Using formula of distance
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=45\times4.60[/tex]
[tex]d=207\ m[/tex]
We need to calculate the maximum height
Using equation of motion
[tex]v^2=u^2-2gh[/tex]
Put the value into the formula
[tex]0=(45.0)^2-2\times9.8\times h[/tex]
[tex]h=\dfrac{(45.0)^2}{2\times9.8}[/tex]
[tex]h=103.3\ m[/tex]
Hence, (a). The distance is 207 m.
(b). The initial velocity is 45.0 m/s
(c). The maximum height is 103.3 m
A runner moves at an average speed of 3.0 m/s and covers 9.5 km. In seconds, how long does the run take?
28.5 s
29.0 s
3.16 s
3200 s
Answer:
3200 seconds
Explanation:
If he ran at 3 m/s for 28.5 seconds, he would have run 85.5 meters
29 seconds would be 87 meters run not 9.5 km
which of the following is the same as 1 second
Answer:
theres no
Explanation:
picture or anything so... ur question doesnt make sense
Why does sonar use ultrasound rather than sound?
Answer; Ultrasonic waves (sounds having frequency greater than 20,000 Hz) are used in sonar because: ... Ultrasonic waves can penetrate water to long distances (because of their high frequency and very short wavelength), but ordinary sound waves or infrasonic waves cannot penetrate water to such long distances.
Answer:
Ultrasonic waves (sounds having frequency greater than 20,000 Hz) are used in sonar because: ... Ultrasonic waves can penetrate water to long distances (because of their high frequency and very short wavelength), but ordinary sound waves or infrasonic waves cannot penetrate water to such long distances
Explanation:
A rocket takes off from Earth. It travels 825km in 75 seconds. What is the
speed of the rocket?
0 1 km/s
11 km/s
O 1.1 km/s
O 111 km/s
Answer:
11 km/s
Explanation:
v=s/t
v=825km/75s
v=11km/s
Which of the following would be best for absorbing light?
a A mirror
b A black shirt
c glass
d concrete