The net force on the cart is 100 N in the direction of its motion, so by Newton's second law we can find the acceleration a applied to it:
100 N = (50 kg) a
a = (100 N) / (50 kg)
a = 2 m/s²
The cart starts at rest and travels a distance of 10 m, so that its final velocity v satisfies
v ² - 0² = 2 (2 m/s²) (10 m)
v ²= 40 m²/s²
and so the cart ends up with kinetic energy
KE = 1/2 m v ² = 1/2 (50 kg) (40 m²/s²) = 1000 J
Which form of energy has the most potential globally
a) Solar energy
b) Bio energy
c) Wind energy
d) Geothermal energy
Answer:
i think the answer is solar,
Explanation:
because it is a natural resource im sorry if im wrong
A steam catapult applies a very large force to supplement the engine thrust of a jet fighter as it takes off from an
aircraft carrier. The jet has a mass of 4000 kg and must reach its take off speed of 70 m/s at the end of the 220
m flight deck. What is the combined force of the catapult and engines? (Use the work-energy theorem)
Answer:
The combined force of the catapult and engines is 44545.455 newtons.
Explanation:
Let suppose that platform of the aircraft carrier is horizontal, such that changes in gravitational potential energy can be neglected. In addition, effects of non-conservative forces are neglected. By using Principle of Energy Conservation and Work-Energy Theorem we find that steam catapult-jet aircraft is represented by:
[tex]K_{1}+W_{1\rightarrow 2} = K_{2}[/tex] (1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies of the jet fighter, measured in joules.
[tex]W_{1\rightarrow 2}[/tex] - Work done on the jet fighter due to the combined force of the catapult and engines, measured in joules.
By applying definitions of translational kinetic energy and work, we expand and simplify the equation above as follows:
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]
[tex]F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot \Delta s}[/tex] (2)
Where:
[tex]F[/tex] - Combined force of the catapult and engines, measured in newtons.
[tex]m[/tex] - Mass of the jet fighter, measured in kilograms.
[tex]\Delta s[/tex] - Length of the catapult, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the jet aircraft, measured in meters per second.
If we know that [tex]m = 4000\,kg[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]v_{2} = 70\,\frac{m}{s}[/tex] and [tex]\Delta s = 220\,m[/tex], then the combined force of the catapult and engines is:
[tex]F = \frac{(4000\,kg)\cdot \left[\left(70\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (220\,m)}[/tex]
[tex]F = 44545.455\,N[/tex]
The combined force of the catapult and engines is 44545.455 newtons.
The combined force of the catapult and engines is 44,545.5 N.
The given parameters;
mass of the jet, = 4000 kgspeed of the jet, v = 70 m/slength of the deck, d = 220 mThe combined force of the catapult and engines is determined by applying work energy theorem as shown below;
[tex]W = \Delta E\\\\Fd = \frac{1}{2}m(v^2 -u^2) \\\\220F = \frac{1}{2}(4000)(70^2 - 0^2) \ \\\\220 F = 9,800,000\\\\F = \frac{9,800,000}{220} \\\\F = 44,545.5 \ N[/tex]
Thus, the combined force of the catapult and engines is 44,545.5 N.
Learn more here:https://brainly.com/question/10063455
a. A remote-control car with a constant velocity drives off the top of a wall that is 10.0 m high and lands 4.60 m from the base of the wall. Draw a diagram of the problem. Label the known and unknown quantities. b. What is the car's speed before it drives off the top of the wall? c. What is the car's vertical speed just before impact? d. How far from the base of the wall would the remote-control car have landed if it had been subject to the gravitational field on the Moon (1.7 N/kg) ?
Answer:
a. Please see the attached graph and drawing combined created with Microsoft Excel
b. Approximately 3.2217 m/s
c. Approximately 14.0 m/s
d. Approximately 11.05 meters
Explanation:
a. The given parameters are;
The height of the wall, h = 10.0 m
The distance from the base of the wall the car lands = 4.60 m
The time, t, it takes the car to land is given by the equation for free fall as follows;
h = 1/2·g·t²
Where;
g = The acceleration due to gravity = 9.81 m/s²
From the equation for free fall, we have;
h/(1/2·g) = t²
∴ t = √(h/(1/2·g) ) = √(10/(1/2·9.81) ) ≈ 1.4278
The time it takes the car to land, t ≈ 1.4278 seconds
b. The horizontal speed of the car = Horizontal distance/(Time) = 4.6/1.4278 ≈ 3.2217 m/s
The horizontal speed of the car before it drives off the wall ≈ 3.2217 m/s
c. The car's vertical speed just before impact is given by the following equation;
v = u + gt
Where;
u = The initial vertical speed = 0
t = The time it takes before impact ≈ 1.4278 seconds
∴ v = 9.81 × 1.4278 ≈ 14.0 m/s
The car's vertical speed just before impact, v ≈ 14.0 m/s
d. Whereby the car is subject to the gravitational field of the moon, we have;
Gravitational force per kilogram = 1.7 N/kg
∴ Gravitational acceleration = 1.7 m/s².
The time it takes the car to land whilst subject to the gravitational field of the moon is therefore;
[tex]t_{Moon}[/tex] = √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43
[tex]t_{Moon}[/tex] ≈ 3.43 seconds
The horizontal distance covered, at the car's horizontal speed in the time of free fall ≈ 3.2217 m/s × 3.43 seconds ≈ 11.05 meters
The horizontal distance covered, at the car's horizontal speed in the time of free fall = The distance the car will land from the base of the wall ≈ 11.05 meters
The distance the car will land from the base of the wall ≈ 11.05 meters.
As the rate of radioactive decay becomes smaller half life become...
pls answer this
quick as possible
Answer:
The answer is the 4th one
Explanation:
Good luck man
A car starts from rest and reaches a velocity of 30 m/s in 10 s. If the car
has a mass of 1500 kg, what net force is exerted upon the car?
Answer:
f= ma
Explanation:
v-u/ t
30-0/10
3m/s
maas 1500
1500*3
4500
Which of the following is NOT a correct statement?*
As the volume of a contained gas increases, the pressure in the container will
decrease.
As the temperature decreases, the kinetic energy of the particles decreases.
As pressure increases with a constant temperature, the volume decreases.
O As temperature increases, the kinetic energy of the particles decreases.
AnswerAmontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. ... If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).:
Explanation:
How much tension must a rope withstand if it is used to accelerate a 1200-kg car vertically upwards at 0.8m/s2?
Answer:
T = 12732 [N]
Explanation:
To solve this problem we must use newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
By means of the attached free body diagram, we can better understand the solution to this problem.
As we see in the diagram the tension Force T lifts the car upwards so the movement is upwards. Therefore we have two forces one upward (positive) of the force T and the other negative downward due to the weight of the vehicle.
As the movement is up the acceleration is also up (with a positive sign).
ΣF = m*a
where:
F = forces [N] (units of Newtons)
m = mass = 1200 [kg]
a = acceleration = 0.8 [m/s²]
T - 1200*9.81 = 1200*(0.8)
T = 12732 [N]
Answer:
12,720
Explanation:
Explain the "war of the currents"
Answer:
the war of currents is
Explanation:
The war of currents was a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s.
Answer:
a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s
Explanation:
During the slowing down process, there is a time when the angular speed is 0.5 rev/s. The fan arm is 0.8 m long. What is the speed of a point on the end of the fan arm at this time
Answer:
2.51 m/s
Explanation:
Given that,
Angular speed of a fan, [tex]\omega=0.5\ rev/s = 3.14\ rad/s[/tex]
Length of the fan arm, r = 0.8 m
We need to find the speed of a point on the end of the fan arm at this time. Let v is the speed. It is given in terms of angular speed is given by :
[tex]v=r\omega\\\\v=0.8\times 3.14\\\\v=2.51\ m/s[/tex]
So, the required speed is 2.51 m/s.
Which action would decrease the strength of attraction between a magnet and an iron object
A. Moving the magnet farther away from the iron object
B. Moving the iron object closer to the magnet
C. Decreasing the number of electrons on the magnet
D. Increasing the number of electrons on the magnet
Answer:
a. moving the magnet farther away and an iron object.
Explanation:
Answer:A
Explanation:
A p e x
Light travels fastest through which of the following?
A.
air
B.
glass
C.
vacuum
D.
water
which property of potential energy distinguishes it from kinetic energy
Answer:
Shape and position
Explanation:
Hope this helps! :)
Need help ASAP please help
Answer:
.....
Explanation:
Why is struggling an important part of the process of learning?
Answer:
Students need a safe environment to take risks and struggle. It’s uncomfortable to struggle, but struggling—falling down and getting back up—is an important facet to learning. Productive struggle is not about being in pain or becoming frustrated.
Please mark branliest!
Explanation: it shows that your really trying your best and when you mess up its telling you that your actually learning from your mistakes.
What type of numbers in a measurement are always significant?
Answer:
Non-zero digits
Explanation:
umm (ꏿ﹏ꏿ;)
Answer:
There's three rules on determining how many important quantities are in a number: Non-zero digits are always predominant. Any zeros between two significant digits are always major. The final zero or dangling zeros in the decimal segment are really only important.
I hope this answered your question, have a nice day!
You tie the loose end of a 0.1 kg yo-yo string to your finger and then release the yo-yo so that it spins down toward the ground (the yo-yo is released from rest and the end of the string tied to your finger remains motionless). After the yo-yo falls a distance of 0.9 m, it has a translational speed of 4 m/s and an angular speed of 180 rad/s. What is the moment of inertia of the yo-yo
Answer:
The answer is "[tex]5.06 \times 10^{-6} \ kg \ m^2[/tex]"
Explanation:
[tex]\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}[/tex]
[tex]= 5.06 \times 10^{-6} \ kg \ m^2[/tex]
Which describes an image that a concave mirror can make?
A. The image is always virtual.
B. The image can be either virtual or real.
C. The image is never upside down.
D. The image can only be the same size as the object.
Answer:
A
Explanation:
The image formed by the concave mirror is always virtual. An enlarged image is caused when the concave mirror is positioned too near to the object.
What is a concave mirror?When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.
A concave mirror is a name for this sort of mirror. An enlarged and fake image is caused when the concave mirror is positioned too near to the object.
When the distance between the item and the mirror is increased, however, the size of the image decreases, and a true picture is generated.
The picture created by the concave mirror may be manipulated. The image formed by the concave mirror is always virtual.
Hence, option A is correct.
To learn more about the concave mirror, refer to the link;
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If a second planet were of the same radius R and made of the same material but had a hollow center of radius 0.50 R , what would be the acceleration of gravity at its surface
Answer:
[tex]g_2=\dfrac{7}{8}g[/tex]
Explanation:
G = Gravitational constant
M = Mass of planet
R = Radius of planet
Acceleration due to gravity on first planet
[tex]g=\dfrac{GM}{R}[/tex]
Assuming that the planets have the same mass density [tex]\rho[/tex]
Density of first planet
[tex]\rho=\dfrac{M}{V}\\\Rightarrow \rho=\dfrac{M}{\dfrac{4}{3}\pi R^3}\\\Rightarrow M=\rho \dfrac{4}{3}\pi R^3[/tex]
Density of second planet
[tex]\rho=\dfrac{M_2}{V_2}=\dfrac{M_2}{\dfrac{4}{3}\pi R^3-\dfrac{4}{3}\pi (0.5R)^3}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{2^3})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{8})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(\dfrac{7}{8})}\\\Rightarrow M_2=\rho\dfrac{4}{3}\pi R^3(\dfrac{7}{8})\\\Rightarrow M_2=M\dfrac{7}{8}[/tex]
Acceleration due to gravity on the second planet
[tex]g_2=\dfrac{GM_2}{R}\\\Rightarrow g_2=\dfrac{GM\dfrac{7}{8}}{R}\\\Rightarrow g_2=\dfrac{7}{8}g[/tex]
The acceleration due to gravity of the planet would be [tex]\dfrac{7}{8}[/tex] times the acceleration due to gravity on the first planet.
The graph below shows the velocity of a car as it attempts to set a speed
record.
Velocity vs. Time
1400
1300
1200
1100
1000
4 45
3
1 (s)
At what point is the car the fastest?
A. t = 1.0 s
B. t = 4.2 s
C. t = 3.0 s
D. t = 4.5 s
From the graph, it is clear that, the velocity is at a time of 1 s is highest. The velocity at 1 second corresponds to 1250 km/hr. Then it decreases with time.
What is velocity - time graph ?The velocity - time graph shows the change in velocity with respect to time. The velocity is placed in y -axis and time is given in x - axis. The slope of the curve in velocity - time graph gives the acceleration of the object.
Similarly, the position of the object in meter after a t seconds can be determined from the velocity - time graph. It is the rate of change in velocity of the object.
From the graph, it is clear that, the curve has its peak at 1 second. After that the peak descends down. Hence, the maximum velocity of the car is at a time of 1 second at which the velocity is 1250 km/hr.
Find more on velocity - time graph :
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At 96° F the saturation density of air is 0.04 kg/m3. If the weather report says the relative humidity is 90 % when the temperature is 96°F, the humidity is
Answer:
The value is [tex]H = 0.036 \ kg/m^3[/tex]
Explanation:
From the question we are told that
The saturation density of air is [tex]\rho = 0.04 \ kg/m^3[/tex]
The temperature for this saturation density to occur [tex]T_s = 96^o F[/tex]
The relative humidity at this temperature is [tex]R_h = 90\% = 0.90[/tex]
Generally the relative humidity is mathematically represented as
[tex]R_H = \frac{H}{\rho}[/tex]
Here H represents humidity
=> [tex]0.90 = \frac{H}{0.04}[/tex]
=> [tex]H = 0.90 * 0.04[/tex]
=> [tex]H = 0.036 \ kg/m^3[/tex]
Which factors affect the rate of deposition? Check all that apply.
color of the sediment
density of the sediment
size of the sediment
texture of the sediment
velocity of the agent of erosion
Answer:
density of the sediment, size of the sediment, and velocity of the agent of erosion.
Explanation:
Have a nice day!
Answer: b., c., e.
Explanation:
asheed and Sofia are riding a merry-go-round that is spinning steadily (uniform circular motion). Sofia is twice as far from the axis as is Rasheed. Sophia's acceleration is _________ that of Rasheed. *
Answer:
The same as
Explanation:
First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.
The angular velocity of a merry go round, in this instance, is given by
w = Δ/Δt
The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.
Transferring this further, the angular acceleration is given as
α = Δw/Δt
Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration
describe the motion of the fireball in the air
Explanation:
doesn't get the question and the answer to
Answer:
(this is the sample answer:After the launch, the fireball moves upward, than it falls back down to EarthEartEarthEarthEarthEarth
Explanation:
What is the average emf between the ends of the wings of a plane flying at a speed of 500 km/hr when the vertical component?
Answer:
This question is incomplete
Explanation:
This question is incomplete because there are some parameters that are missing. However, when calculating the average emf and speed/velocity is provided, then
average emf = Brvl
where Br is the vertical component of the earth's magnetic field (in weber per square meter, Wb/m²)
v is velocity (in m/s)
l is the length (in meter) of the wing span
1500 POINTS TO WHOEVER ANSWERS THIS FIRST!!!!!!!!!!
1. You drop a bowling ball 15m from the top of the science building. What is its
velocity as it impacts the ground?
2. A cheetah accelerates from 0 to 13m/s in 3 seconds. How far does it travel
during that time?
3. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. How high does the bullet go (neglecting air resistance)?
4. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. What is the bullets velocity when it falls back down and hits your
friend?
Answer: Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.
Explanation:
what can you infer from the statement. velocity is an object is zero
Which of the following is how the substance in a longitudinal wave flows?
A. Up and Down
B. At Random
C. Perpendicular to the direction of the wave
D. In the direction of the wave
Please help I need this turned in tonight.
Answer:
The answer is D. In the direction of the wave
Explanation:
I took the test, hope this helps! :)
What is the momentum of 100 kg running at 4m/s north?
Bicyclist travels at an average velocity of 11.2 km/h [W]. How far will the bicyclist travel in 175 minutes?
Answer:
32.67km
Explanation:
~= round off
11.2km/h~3.11m/s
175minutes=10500s
10500*3.11~32, 666.67m
=32.66667km
~ 32.67km