By Newton's second law, the car has acceleration a such that
2000 N = (1500 kg) a
a = (2000 N) / (1500 kg)
a = 4/3 m/s² ≈ 1.33 m/s²
After 15 s, the car will attain a speed of
(4/3 m/s²) (15 s) = 20 m/s
It will have traveled a distance ∆x such that
(20 m/s)² - 0² = 2 (4/3 m/s²) ∆x
∆x = (20 m/s)² / (8/3 m/s²)
∆x = 150 m
How much Tension force is required to pull a 1500 kg car (it is being towed) forward with an acceleration of 3 m/s^2 if the friction force on the towed car's tires is pulling backward with a force of 2500 N?
If the pull is done horizontally, then the net force on the car is
∑ F = T - f = (1500 kg) (3 m/s²)
where T is the magnitude of the tension in the towing cable, and f is the friction which points in the opposite direction. Then
T = f + (1500 kg) (3 m/s²)
T = 2500 N + 4500 N
T = 7000 N
Please help
Which point has the most Potential energy? [Select]
Which point has the most Kinetic energy? [ Select ]
Answer:
Most potential energy: A
Most kinetic energy: D
Explanation:
Kinetic Energy is the type of energy an object has due to its state of motion. It's proportional to the square of the speed.
The equation for the kinetic energy is:
[tex]\displaystyle K=\frac{1}{2}mv^2[/tex]
Where:
m = mass of the object
v = speed at which the object moves
The gravitational potential energy is the energy stored in an object because of its height h in a gravitational field.
It can be calculated with the equation:
U=m.g.h
The point where the object has the most potential energy is that where it has more height. It corresponds to point A.
When the object is at zero height, all of its potential energy was transformed to kinetic, thus the point where the kinetic energy is D.
Most potential energy: A
Most kinetic energy: D
Answer:
potentail enegry at point a and kinetic enegy at point c
Explanation:
i did this i fourth grade please mark brainlist
4. A Ferrari travels 350km in 2 hours. What was it's
s speed?
Could somebody please explain the Coriolis effect? Thank you! (also the subject is physical science but that wasn't an option so I just put physics)
Answer:
The Coriolis Effect makes things appear to move in a curve around our planet, like, for example, a plane. This is also due to the fact that our Earth is round, so nothing really moves in a straight line. It is "an effect whereby a mass moving in a rotating system experiences a force (the Coriolis force ) acting perpendicular to the direction of motion and to the axis of rotation."
Explanation:
A stretched spring attached to two fixed points is compressed on one end and released, as shown. The resulting wave travels back and forth between the two fixed ends of the spring until it comes to a stop. This is an example of _____ .
a Transverse wave
b
longitudinal wave
c
surface wave
d
electromagnetic wave
Answer:
B. Longitudinal wave
Explanation:
It's the only option that makes sense. Hope this helped :)
A car accelerates from rest at a constant acceleration of 25.0 m/s^2. At some point, it then turns off its engine, letting the car decelerate slowly from the force of friction at a constant deceleration of 3 m/s^2 until it is at rest again. The total speed the car moves in this time is 200 meters. What is the minimum time needed for the car to move 200 meters given that it both starts and ends at rest?
Answer:
t = 9.14 s
Explanation:
We first analyze the accelerating motion by applying first equation of motion:
Vf₁ = Vi₁ + a₁t₁
where,
Vf₁ = Final Speed of Car before turning off engine
Vi₁ = Initial Speed of Car = 0 m/s
a₁ = acceleration of car = 25 m/s²
t₁ = time taken in accelerating motion
Therefore,
Vf₁ = 25t₁ ---------- equation (1)
Now, we apply second equation of motion:
s₁ = Vi₁ t₁ + (1/2)a₁t₁²
where,
s₁ = distance covered during accelerating motion
Therefore,
s₁ = (0)t₁ + (1/2)(25)t₁²
s₁ = 12.5 t₁² ----------- equation (2)
Now, we analyze the decelerating motion by applying first equation of motion:
Vf₂ = Vi₂ + a₂t₂
where,
Vf₂ = Final Speed of Car = 0 m/s
Vi₂ = Initial Speed of Car after turning off engine
a₂ = deceleration of car = - 3 m/s²
t₂ = time taken in decelerating motion
Therefore,
Vi₂ = 3t₂ ---------- equation (3)
Now, we apply second equation of motion:
s₂ = Vi₂ t₂ + (1/2)a₂t₂²
where,
s₂ = distance covered during decelerating motion
Therefore,
s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²
s₂ = Vi₂ t₂ - 1.5 t₂²
using equation (3):
s₂ = 3 t₂² - 1.5 t₂²
s₂ = 1.5 t₂² ------------ equation (4)
Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):
Vf₁ = Vi₂
using equation (1) and equation (3):
25 t₁ = 3 t₂
t₁ = 0.12 t₂ ------------ equation (5)
Also, we know that sum of the distances is 200 m:
s₁ + s₂ = 200
using equation (2) and equation (4):
12.5 t₁² + 1.5 t₂² = 200
using equation (5):
12.5 (0.12 t₂²) + 1.5 t₂² = 200
3 t₂² = 200
t₂² = 200/3
t₂ = 8.16 s
substitute this in equation (5):
t₁ = 0.12(8.16 s)
t₁ = 0.97 s
Hence, the minimum time required for this motion is:
t = t₁ + t₂ = 0.97 s + 8.16 s
t = 9.14 s
As we know, the moon is a satellite of our earth, what is the
theoretical period of the moon? The average radius of the
moon's orbit is 3.84 108 m and the mass of the earth is 5.97 x
1024 kg (in hours, G = 6.67 x 10-9 N (m/kg) 3).
Answer:c
Explanation:c
This question involves the concepts of the time period, orbital radius, and gravitational constant.
The theoretical period of the moon is "658 hr".
The theoretical time period of the moon around the earth can be found using the following formula:
[tex]\frac{T^2}{R^3}=\frac{4\pi^2}{GM}[/tex]
where,
T = Time Period of Moon = ?
R = Orbital Radius = 3.84 x 10⁸ m
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
Therefore,
[tex]\frac{T^2}{(3.84\ x\ 10^8\ m)^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\T^2=(9.91\ x\ 10^{-14}\ s^2/m^3)(56.62\ x\ 10^{24}\ m^3)\\\\T=\sqrt{561.34\ x\ 10^{10}\ s^2}[/tex]
T = 2.37 x 10⁶ s[tex](\frac{1\ h}{3600\ s})[/tex]
T = 658 hr
Learn more about the orbital time period here:
https://brainly.com/question/14494804?referrer=searchResults
The attached picture shows the derivation of the formula for orbital speed.
3. A cart with mass of 30 kg is traveling with a velocity of 4.0 m/s. The
cart then gains speed, achieving a new velocity of 10 m/s after 4
seconds. Calculate the magnitude of the net force acting on the cart.
Answer:
39 m/s
Explanation:
Answer:
39 m/s
Explanation:
A 55 kg boy running at 2.0 m/s jumps onto a 2.0 kg skateboard. Calculate the final velocity of the boy and the skateboard.
Answer:
The final velocity of the boy and the skateboard is 1.93 m/s.
Explanation:
We can find the final velocity of the boy and the skateboard by conservation of linear momentum:
[tex] P_{1} = P_{2} [/tex]
[tex] m_{b}v_{i_{b}} + m_{s}v_{i_{s}} = m_{b}v_{f_{b}} + m_{s}v_{f_{s}} [/tex] (1)
Where:
[tex]m_{b}[/tex]: is the boy's mass = 55 kg
[tex]m_{s}[/tex]: is the skateboard's mass = 2.0 kg
[tex]v_{i_{b}}[/tex]: is the initial speed of the boy = 2.0 m/s
[tex]v_{i_{s}}[/tex]: is the initial speed of the skateboard = 0
[tex]v_{f_{b}}[/tex]: is the final speed of the boy =?
[tex]v_{f_{s}}[/tex]: is the final speed of the skateboard =?
Since the boy and the skateboard will have the same final velocity by entering the above values into equation (1) we have:
[tex] 55 kg*2.0 m/s + 0 = v_{f}(55 kg + 2.0 kg) [/tex]
[tex] v_{f} = 1.93 m/s [/tex]
Therefore, the final velocity of the boy and the skateboard is 1.93 m/s.
I hope it helps you!
The final velocity of the boy and the skateboard is 1.93m/s
According to the law of conservation of momentum, the momentum of the bodies before the collision is equal to the momentum of the bodies after the collision.
Mathematically, [tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
[tex]m_1 \ and \ m_2[/tex] are the masses of the object
[tex]u_1 \ and \ u_2[/tex] are the velocities of the object
v is the final velocity
Given the following parameters
m₁ = 55kg
u₁ = 2.0m/s
m₂ = 2.0kg
u₂ = 0m/s
Required
Final velocity "v"
Substitute the given parameters into the formula to have:
[tex]55(2)+2(0)=(55+2)v\\110=57v\\v=\frac{110}{57}\\v= 1.93m/s[/tex]
Hence the final velocity of the boy and the skateboard is 1.93m/s
Learn more here: https://brainly.com/question/22257327
One of the major differences between our common Physics models of energy change and realistic models of them is
In the Physics models, the acceleration due to gravity is rounded to a non-exact but easier to use number
In the Physics models, the velocities are only calculated at set intervals of time, instead of continuously
In the Physics models, the objects are assumed to have simplified shapes in order to make motion by acceleration easier to calculate
In the Physics models, we do not take into account the energy change by the friction of moving components
Answer: In the Physics models, we do not take into account the energy change by the friction of moving components
Explanation:
INEGRATED SCIENCE-8th grade science
report cards due dec.18
Please Help ASSSAAAAPPPP! PLEASEEEEEEE
Answer:
I do drink bottled water but not to often, I live about 2.8 miles away from its water source and if i did'nt drink bottled water and I had to found out how my drinking water is treated I would go ask the people that are taking care of the water.
Explanation:
hope this helps
A river flows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 10.0 m/s due north relative to the water. If the river is 325 m wide, how far downstream is the boat when it reaches the north shore?
Answer:
Explanation:
Given that the velocity of the river is 1.50m/s due east.
The velocity of the boat with respect to the river is 10.0 m/s due north .
The width of the river is 325 m.
Note that the velocity of the boat will help in crossing the river.
Hence, speed = distance/time
Speed = 10.0 m/s
Distance = 325 m
Time = ?
10 = 325/time
Time = 325/10
Time = 32.5s
Hence in 32.5s the horizontal distance traveled by boat will be
= 1.50m/s × 32.5s
= 48.75m
does an object in motion stay in motion
; use three examples of how dance can be used in somebody creating a fitness plan
1 class comment
Answer: Physical fitness can be defined as the ability of a person to perform daily routine tasks without any ailment.
Explanation:
Dance workouts can help in improving the physical fitness of the person.
This may involve improvement of the cardiovascular strength.
This may also help in improving the strength and endurance of a person.
This improves the stamina, flexibility, and composition of the body.
The dance also helps in maintaining balance and attaining the correct posture.
Integrated Science- 8th grade science
Please Help ASAP!
Its almost the end of semester And I need to fix my grade for science so PLEASE help!
its almost report cards time!
Answer:
A. a rigorously tested explanation
Explanation:
B. and D. are out - theories are not opinionated, they are factualC. is out - not all theories are mathematicalA. is the best choiceAnswer:
The answer is option letter B
what makes up a atom
Answer:
They're typically made up of three main parts: protons, neutrons and electrons. Think of the protons and neutrons as together forming a “sun”, or nucleus, at the centre of the system. The electrons orbit this nucleus, like planets. If atoms are impossibly small, these subatomic particles are even more so.
Explanation:
hope i helped.
Answer:
Atoms consist of a nucleus made of protons and neutrons orbited by electrons. ... We now know that atoms are made up of three particles: protons, neutrons and electrons — which are composed of even smaller particles, such as quarks.
Explanation:
In which of the following situations would there be initial energy in the system?
An apple sits motionless on the ground near a tree
A spring is compressed and held at compression by a person before releasing it
A bowstring is neither pulled nor stretched on a bow
An object rests at a velocity of 0 on a flat surface at ground level
Answer:
A spring is compressed and held at compression by a person before releasing it
Explanation:
All the other answers are showing things that are at rest and have no energy starting nor going through them. This answer show how the energy is starting and being released.
The largest acceleration that a human has ever endured occurred when a race car accidentally crashed into a wall. The car was traveling at a speed of 172.8 km/h when it hit the wall. The car came to a complete stop 2.72 ´ 10–2 s later.
a. Calculate the acceleration of the car using the acceleration formula. Express your answer in both m/s2 and in “g’s.” One g is equal to the free-fall acceleration of 9.8 m/s2.
b. Suppose the driver of the car had a mass of 70.0 kg. What was the unbalanced force on his body as the car underwent negative acceleration?
Answer:
(a) The acceleration oF the car is [tex]-1764.7 m/s^2[/tex] or -180.1 g’s.
(b) The magnitude of the force on the driver's body is 123529 Newtons
Explanation:
Given that:
The initial velocity of the car before the impact, u=172.8 km/h
Or, [tex]u =172.8 \times \frac {1000}{3600} m/s= 48m/s[/tex]
As the car came to a complete stop after the collision, so the final velocity of the car, v=0.
The time required to stop the car completely, [tex]t=2.72\times 10^{-2} s[/tex]
(a) Assuming the acceleration is constant, so
Acceleration, [tex]a= (v-u)/t[/tex]
So, [tex]a=\frac {0-48}{2.72\times 10^{-2}}=-1764.7 m/s^2[/tex]
Now, expressing the acceleration in the unit od "g's",
[tex]a=-1764.7\times \frac{g}{g} \\\\\Rightarrow a=-1764.7 \;m/s^2\times \frac{g}{9.8\; m/s^2} \\\\\Rightarrow a=-180.1 g[/tex]
Hence, the acceleration od the car is -1764.7 [tex]m/s^2[/tex] or -180.1 g’s.
(b) Mass of the driver of the car, m=70.0 kg
The acceleration of the body of the driver is the same as the acceleration of the car.
So, the acceleration faced by the driver, [tex]a= -1764.7 m/s^2[/tex]
As forec = mass x acceleration, so
The force on the driver's body = 70 x (-1764.7)= -123529 Newtons.
The negative sign means the force is in the opposite direction of the direction of motion of the car.
Hence, the magnitude of the force on the driver's body is 123529 Newtons
PLEASE HELP ME IM TIMED
Answer:
the answer is the core
Explanation:
the core is composed of iron and nickel
How much power is used if a force of 35 newtons is used to push a box a distance of 10 meters in 5 seconds?w=350j
Answer:
How much power is used if 350J of work is done when pushing a box for 5 seconds.
Explanation:
the answer is 70watts
Students are asked to design an experiment about Newton’s 2nd Law. One student decides to roll a marble down a ramp into a pile of sand to measure the force impact.
Which variable should she manipulate to best exemplify the relationship explained by this law?
A.She should use a heavier marble, because the marbles will roll at the same rate of acceleration but more mass will produce a larger impact force.
B.She should increase the slope of the ramp by propping it up to higher height, because a steeper ramp will cause a greater rate of acceleration and a larger impact force.
C.She should use a heavier marble, because a bigger marble will accelerate more quickly down the ramp and cause a greater impact force.
D.She should decrease the slope of the ramp, because a ramp with a smaller slope will allow the ball more time to build up speed and cause a greater impact force.
Answer:
A
Explanation:
Trust me I just took it !
Question 5 (1 point)
A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the
ground. How far, in meters, from the deck does the ball land on the ground?
Answer:
The horizontal distance the ball travels is approximately 4.055 meters
Explanation:
The given parameters are;
The height from which the child kicks the ball = 3.5 m
The horizontal speed of the ball = 4.8 m/s
Therefore, we have;
The time it takes the ball to hit the ground is given by the relation;
h = u·t + 1/2·g·t²
Where;
u = The initial vertical velocity of the ball = 0 m/s
t = The time it takes the ball to hit the ground
g = The acceleration due to gravity = 9.81 m/s²
h = The height of the ball = 3.5 m
3.5 = 0 × t + 1/2 × 9.81 × t²
3.5 = 1/2 × 9.81 × t²
∴ t² = 3.5/(1/2 × 9.81)
∴ t = √(3.5/(1/2 × 9.81) = 0.8447 s
t ≈ 0.8447 s
The time the ball takes in flight = t ≈ 0.8447 s
Therefore;
The horizontal distance the ball travels = The horizontal velocity × The time of flight
∴ The horizontal distance the ball travels = 4.8 × 0.8447 ≈ 4.055
The horizontal distance the ball travels ≈ 4.055 meters.
A 125,000 kg locomotive is traveling south at 1.0 m/s through a switching yard. A connected set of parked gondola cars weighing 1,750,000 kg are directly ahead. Determine the speed and direction of the completed train after the locomotive has been coupled to the cars.
Answer:
The speed of the completed train is 0.0[tex]\bar 6[/tex] m/s
The direction of the completed train is South
Explanation:
The given parameters are;
The mass of the locomotive, m₁ = 125,000 kg
The initial speed of the locomotive, v₁ = 1.0 m/s
The mass of the gondola cars, m₂ = 1,750,000 kg
The initial sped of the gondola cars, v₂ = 0
Let v₃ represents the speed of the completed train. From the principle of conservation of linear momentum, we have;
m₁·v₁ + m₂·v₂ = (m₁ + m₂)×v₃
Substituting the known values, gives;
125,000 × 1.0 + 1,750,000 × 0 = (125,000 + 1,750,000) × v₃
∴ v₃ = (125,000 × 1.0 + 1,750,000 × 0)/(125,000 + 1,750,000) = 1/15 = 0.0[tex]\bar 6[/tex]
v₃ = 0.0[tex]\bar 6[/tex] m/s
The speed of the completed train = 0.0[tex]\bar 6[/tex] m/s
The direction of the completed train = The direction of the locomotive = South.
A box is being pulled across a horizontal surface by a 20 N force to the right.
a. If the box moves at a constant velocity, what do you know about the forces acting on the object?
b. If the box moves at a constant velocity, how much force opposes the motion of the box?
c. If the box experiences a force of 15 N to the left, along with the 20 N force acting to the right, what is the net force on the box?
A) If the box moves at a constant velocity, what we know about the forces acting on the object is; the sum of the forces acting on the object must be zero.
B) If the box moves at a constant velocity, the force that opposes the motion of the box is; Frictional Force of -20 N
C) If the box experiences a force of 15 N to the left, along with the 20 N force acting to the right, the net force on the box is; F_net = 5 N
We are told that a box is being pulled across a horizontal surface by a 20N force to the right.
A) Newton's first law of motion states that an object will remain at rest or continue in motion unless an external force acts on it. This means that any object/body moving with constant velocity will have no net external force acting on it. Thus, the sum of the forces acting on the object must be zero. B) Since we are told that the box is moving at a constant velocity and that the sum of all forces must be zero. Then the force that will oppose the applied force of 20N to the right will be an equal but opposite force .In this case the opposing force will have to be the frictional force which will be -20 N to balance out the applied force.
C) We are told that a force of 15 N is acting to the left along with the 20 N force.Thus, the net force will be;
F_net = 20 - 15
F_net = 5 N
Read more at; https://brainly.com/question/20712983
Can someone help me in this please any one good in science.
Let's calculate the equivalent resistances on both circuits.
On Diagram A we have a series connection of the resistors. The equivalent resistance will be the sum of all resistances:
[tex]R_{eq}=1+1+1\\\\\boxed{R_{eq}=3\Omega}[/tex]
On diagram B we have a parallel connection of the resistors. The reciprocal of the equivalent resistance will be the sum of the reciprocals of all the resistances:
[tex]\frac{1}{R_{eq}} = \frac{1}{1} +\frac{1}{1} +\frac{1}{1} \\\frac{1}{R_{eq}}=3\\\\\boxed{R_{eq}=\frac{1}{3}}[/tex]
Therefore, the larger resistance occurs on diagram A.
For the current, recall
[tex]V=IR[/tex]
Where [tex]I[/tex] stands for current [tex]R[/tex] is the resistance and [tex]V[/tex] is the voltage. Rearranging the equation we have
[tex]I = \frac{V}{R}[/tex]
We can see that the larger the resistance, the smaller the current gets. So the larger current must happen in the diagram with smaller resistance. Therefore, the larger current occurs on diagram B.
Glad to help, wish you great studies ;)
Mark brainliest if you deem the answer worthy
How do we obtain data about the properties of exoplanets and objects in our solar system?
Answer:
Doppler effect to analyze the motion and properties of the star and planet.
Explanation:
The radial velocity method invovles watching the spectral lines of a star as a planet orbits around the star. Because of the planets gravitational pull on the star, will wobble. This causes Doppler shifts in the spectral lines, allowing astronomers to infer the presence of the planets.
1. A large ocean liner floating in the sea has a volume of
375 000 m² and displaces 50 000 m² of sea water. Determine the
density and mass of the ship. Explain why, despite being made
of metal, the ship is able to float.
Answer:
Density of the ship :133.33 kg/m³
Mass of the ship :50,000,000 kg
Ships are able to float on water because they displace a volume that equals their own weight
Explanation:
The formula for density is ;
Density = Mass/Volume
But the mass of displaced fluid is give by the formula;
m=ρV where ρ is density of the fluid and V is volume of water displaced
Volume of liner= 375000 m³
Volume displaced = 50000 m³
The volume of the water displaced is equal to mass of the liner
I m³ = 1000 kg
50000 m³ = 50000*1000 =50,000,000 kg
50,000,000 kg ----------mass of the ship
So if ;
m= 50,000,000 kg
v= 375000 m³
ρ = m/V
ρ = 50,000,000/375000
ρ = 50,000,000/375000
ρ =133.33 kg/m³
Ships are able to float on water because they displace a volume that equals their own weight. Additionally, the air contained in the spaces of the ship inside is much less dense than density of water aiding in the floating process.
An object is in projectile motion if it?
Answer:
A projectile is an object upon which the only force is gravity. ... The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity. Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity.
true or false? drag forces decrease when the density of a fluid increases
Answer:
Drag increases with the density of the fluid (ρ). More density means more mass, which means more inertia, which means more resistance to getting out of the way. The two quantities are directly proportional.