Answer:
The answer is "4.97 m".
Explanation:
[tex]u = 1.23\ \frac{m}{s}\\\\[/tex]
[tex]H= 1.45 \ m\\\\[/tex]
[tex]\mu = 0.231\\\\[/tex]
The law of conservation tells us that heat energy at the top with kinetic energy at the top equals kinetic energy at the base.
[tex]mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\\\2gh +u^2 =v^2\\\\v=\sqrt{u^2+2gh}[/tex]
[tex]v=\sqrt{(1.23\ \frac{m}{s})^2+2(9.81 \frac{m}{s^2}) +(1.45\ m)[/tex]
[tex]=\sqrt{1.5129+19.62 +1.45}\\\\=\sqrt{22.5829}\\\\=4.75\ \frac{m}{s}[/tex]
Friction force is given by the formula
[tex]f=-\mu mg \\\\ma= -\mu mg\\\\a=-\mu g\\\\[/tex]
[tex]= -(0.231) \ (9.81\ \frac{m}{s^2})\\\\=-2.26611 \ \frac{m}{s^2}[/tex]
Now by using an equation of motion as
[tex]v^2-u^2= 2as[/tex]
From the above the distance traveled is
[tex]S=\frac{v^2-u^2}{2a}[/tex]
[tex]S=\frac{(0)^2-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\[/tex]
[tex]=\frac{-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\=\frac{-22.5625}{-4.53222}\\\\=4.97[/tex]
In other words, the distance from the bottom of the chute to the point where the toy zebra comes to rest is [tex]s = 4.97\ m[/tex]
i. a A ball is released from a height of 45 m top of the building. If it strikes the ground surface after 5 seconds, calculate the acceleration and final velocity of the ball. [18 m/s, 3.6 m/s²].
Answer:
3.6m/s², 18m/s
Explanation:
So according to h=1/2at², 45=1/2a*25, a=3.6m/s² (which means that it didn't happen on Earth?)
Final velocity = at = 3.6*5 = 18 m/s
define the unit of work
[tex]\text{The unit of work is same as energy. Newton-meters(Nm) or Joules(J).}\\[/tex]
A plane flies from New York to LA at a constant speed of 800 km an hour how long will it take the plane to fly to 4,200 kilometers
Answer:
3.5 hours
Explanation:
A 3-column table with 1 row. The first column titled distance travelled (meters) has entry 6. 1. The second column labeled lower track elapsed time (seconds) has entry 4. 92. The third column labeled higher track elapsed time (seconds) has entry 3. 36. Based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the higher track? How can the reasoning for the above answer be best explained? On the higher track, the elapsed time is. Calculate speeds for each track. How much faster was the car on the higher track than the lower track?.
Answer:
B,A,A
Explanation:
Answer:
Other guy is correct b,a,a
Explanation:
Please help if you can!
Answer:
a
Explanation:
bc i said so its right lol
You plug your microwave into an outlet and then you heat up a piece of pizza in it. This is an example of an...
Microwave Radiation????????????
What determines the amount of gravitational force between
objects?
Answer:
The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them. This means that the force of gravity increases with mass, but decreases with increasing distance between objects.
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what is the affect of density of air in the velocity of sound??
Answer: Hello!
The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to m, the mass of the oscillating object. The speed of sound in air is low, because air is easily compressible.
Explanation:
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A 1.5 kg ball has a velocity of 12 m/s just before it strikes the floor. Find the impulse on the ball if the ball bounces up with a velocity of 10 m/s.
Hi there!
Recall:
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Let the direction TOWARDS the floor be POSITIVE, and AWAY be NEGATIVE.
Plug in the givan values:
Δp = 1.5(-10 - 12) = -33 Ns
**OR, the magnitude: |-33| = 33 Ns
A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in energy of the system
Explanation:
For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.
ΔU=Q−W
We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.
W=FΔx
W=3N×2m
W=6J
Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.
ΔU=Q−W
ΔU=10J−6J
ΔU=4J
Answer is 4J
i think this may help you very much
If a 75 kg box collides with a stationary 35 kg box with a force of 110 N, what must be true of the magnitude of the reaction force?
According to Newton third law of motion, what must be true of the magnitude of the reaction force is that the reaction force will be equal and opposite to the action force of 110 newtons
From Newton third law of motion which state that in every action of force, there will be equal and opposite reaction.
If a 75 kg box collides with a stationary 35 kg box with a force of 110 N, the mass 35 kg will produce a reaction force which is equal and opposite to the action force 110 newton received from mass 75 kg.
Therefore, what must be true of the magnitude of the reaction force from mass 35 kg is that the mass will produce an equal and opposite force equal to 110 Newtons.
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How did Thomson's model get its name?
Answer:
thats actually a really good question. sadly i don't know the answer to it. but im sure that there are others who can!
A car which is traveling at a velocity of 27 m/s undergoes an acceleration of 5.5 m/s^2 over a distance of 430 m. How fast is it going after that acceleration?
73.9m/s (1dp)
1) list everything that you are given using suvat, where s is distance, u is initial velocity(speed), v is final velocity(speed), a is acceleration and t is time
s = 430m
u = 27 m/s
v = ? (we need to work out)
a = 5.5m/s^2
t = (we are not given this value)
2) use an equation that doesn't involve the time
[tex] {v}^{2} = u {}^{2} + 2as[/tex]
3) input the values that we have
[tex] {v}^{2} = ( {27})^{2} + 2(5.5)(430)[/tex]
[tex]v {}^{2} = 5459[/tex]
[tex]v = \sqrt{5459} = 73.9[/tex]
so the answer is 73.9m/s to 1dp
Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 10 m/s.
Hi there!
Kinetic energy can be calculated using the following:
[tex]\large\boxed{KE = \frac{1}{2}mv^2}}[/tex]
Where:
KE = Kinetic energy (J)
m = mass (kg)
v = velocity (m/s)
Plug in the given values:
[tex]KE = \frac{1}{2}(625)(10^2) = \boxed{31250J}[/tex]
1 point
A hairdryer uses 10 A of current when plugged into a 120 V outlet. How
much power does it use?
Answer:
1200 watt
Explanation:
P= I*V=10*120=1200watt
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32. 4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate? The wave has traveled 32. 4 cm in 3 seconds. The wave has traveled 32. 4 cm in 9 seconds. The wave has traveled 97. 2 cm in 3 seconds. The wave has traveled 97. 2 cm in 1 second.
The true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.
In Physics, we define a wave as a disturbance along a medium that transfers energy. The wavelength of a wave is the distance covered by the wave while the frequency of the wave is the number of cycles of the wave completed per second.
The period of the wave is the inverse of the frequency of the wave. It is defined as the time taken for the wave to complete a cycle and it is measured in seconds.
The wave formula is given as;
v = λf
v = velocity of the wave (distance traveled by the wave in one second)
λ = wavelength of the wave
f = frequency of the wave
So;
λ = 32.4 cm
f = 3 hertz
v = 32.4 cm × 3 hertz
v = 97. 2 cms-1
Hence, the true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.
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Find the area of the given figure
Answer:
77 sq.cm
Explanation:
Solution,
Trapezium
Perpendicular (P1)=8cm
Perpendicular(P2)=14cm
height(h)=7cm
Now,
We know that,
Area(A)= 1/2 ×h (P1+P2)
= 1/2 × 7cm (8cm+14cm)
= 1/2 × 7cm(22cm)
= 1/2 × 154 squarecm
=77 sq.cm
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
What is the efficiency of the lamp?
С
18%
D
85%
A
0.15%
15%
B
Answer:
[tex]15\%[/tex].
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is [tex]40\; \rm W[/tex] (same as [tex]40\; \rm J \cdot s^{-1}[/tex],) meaning that [tex]40\; \rm J[/tex] of energy is supplied to this lamp every second.
The question states that [tex]34\; \rm J[/tex] out of that [tex]40\; \rm J[/tex] of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the [tex](40\; \rm J - 34\; \rm J) = 6\; \rm J[/tex] of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive [tex]40\; \rm J[/tex] of energy input and would outputs [tex]6\; \rm J[/tex] of useful work. The efficiency of this lamp would be:
[tex]\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}[/tex].
A 300 g ball swings in a vertical circle at the end of a 1.3-m-long string. When the ball is at the bottom of the circle, the tension in the string is 13 N.
What is the speed of the ball at that point?
Answer:
0.23N for the speed
at the bottom of the circle
The diagram shows a nephron.
A nephron. W points to branch of renal artery. X points to branch of renal vein. Y points to tubular point of nephron. Z points to collecting duct.
Where is the blood first filtered?
Answer:
w
Explanation:
renal artery...the blood flow into the kidney via this blood vassles
and the filtration take place is called ultrafiltration
The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?(1 point)
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface. How does the velocity of Satellite A compare to the velocity of Satellite B?(1 point)
A) It depends on the masses of the satellites.
B) The velocity of Satellite A is greater than the velocity of Satellite B.
C) The velocity of Satellite B is equal to the velocity of Satellite A.
D) The velocity of Satellite A is less than the velocity of Satellite B.
What is the direction of the net force acting on a satellite as it orbits Earth at a constant speed?(1 point)
A) in the direction the satellite is moving
B) toward the center of Earth
C) away from the center of Earth
D) opposite the direction the satellite is moving
What happens to the gravitational force and orbital velocity of a satellite as the satellite transfers to an orbit that is closer to Earth?(1 point)
A) The gravitational force decreases and the velocity increases.
B) The gravitational force increases and the velocity increases.
C) The gravitational force decreases and the velocity decreases.
D) The gravitational force increases and the velocity decreases.
Newton's second law and universal gravitation law allow to find the results for questions about the motion of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
Newton's second law establishes a relationship between force, mass, and the acceleration of bodies.
F = m a
In the case of the satellite the force is given by the law of universal gravitation.
[tex]F = - G \frac{Mm}{r^2 }[/tex]
Where G is the constant of universal gravitation. M and m the mass of each object and r the distance between them.
In this case the satellite is in a circular orbit, therefore the acceleration is centripetal.
[tex]a = \frac{v^2}{r}[/tex]
We substitute.
[tex]G \frac{Mm}{r^2} = m \frac{v^2}{r}[/tex]
[tex]\frac{GM}{r} = v^2[/tex]
Let's analyze the answers to find the correct one.
1) They indicate the height of the space station r = 4.08 10⁵ m and ask the speed.
[tex]v= \sqrt{ \frac{6.67 \ 10 ^{-11} 5.9 \ 10^{24}}{4.08 \ 10^6 } }[/tex]
v = 9.82 10³ m / s
The correct answer is D.
2) Satellite A has an orbit of hₐ = 500 km and satellite b an orbit of
h_b = 800 km
The distance from the center of the earth to each satellite is:
rₐ = R + hₐ
r_b = R + h_b
rₐ = 6.37 106 + 500 10³ = 6.87 10⁶ m
r_b = 6.37 10⁶ + 800 10³ = 7.17 10⁶ m
Let's find the ratio of the speeds
[tex]\frac{v_a}{v_b} = \sqrt{ \frac{r_b}{r_a} } \\\frac{v_a}{v_b} = \sqrt{ \frac{7.17}{6.87} }[/tex]
[tex]\frac{v_a}{v_b}[/tex] = 1,022
we see that the speed of satellite a is slightly greater than the speed of satellite b.
Let's analyze the claims.
A) False. The speed does not depend on the mass of the satellites.
B) True. The velocity of a is slightly greater than the velocity of b.
C) False. The speed of a is greater.
D) False. The speed of a is greater.
3) As the orbit is circular, the force must be radial, that is, it points towards the center of the earth.
Let's analyze the claims.
A) False. The speed modulus does not change, therefore there is no acceleration in the direction of the satellite.
B) True. Aim for the center of the Earth, change the direction of the velocity.
C) False. Aim for the scepter of the earth.
D) false. The modulus of velocity is constant and the direction changes towards the center of the earth, therefore the force must go towards the center of the earth.
4) The force in the law of universal gravitation increased as the distance decreased.
When a satellite approaches the earth its speed must increase since the speed is proportional in inverse of the square root of the distance.
Let's examine the claims.
A) False. The attraction force increases.
B) True. You agree with the explanation.
C) False. The gravitational force increases.
D) False. Speed increases.
In conclusion, using Newton's second law and the universal law of gravitation we can find the results for the questions about the movement of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
Learn more about the law of universal gravitation and circular motion here: brainly.com/question/24851258
Answer:
1. 7.66 × 10^3 m/s
2. The velocity of Satellite A is greater than the velocity of Satellite B.
3. toward the center of Earth
4. The gravitational force increases and the velocity increases.
Explanation:
4) What is the kinetic energy of o 5 kg object moving at a velocity of 25 m/s?
Ek = 1/2 mv^2
= 1/2 × 5 × (25)^2
= 1562.5 J
draw the following vector quantity Using the coordinate system.
a. 190 newton east
b. 120km/hr, 250 north of east
c. 60 meters southwest
The given vectors quantities can be described by their properties of both
magnitude and direction.
a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.Reasons:
a. The magnitude of the vector = 190 N
The direction in which the vector acts = East
Therefore, in vector form, we have;
[tex]\vec{F}[/tex] = 190 × cos(0)·i + 190 × sin(0)·j = 190·i
The vector can be represented by an horizontal line, 190 units long
Coordinate points on the vector = (0, 0) and (190, 0)
The drawing of the vector with the above points using MS Excel is attached.
b. Magnitude of the velocity vector = 120 km/hr. 25° North of east
Solution;
The vector form of the velocity is; [tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j, which gives;
[tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j
[tex]\vec{v}[/tex] ≈ 108.76·i + 50.714·j
Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)
The drawing of the vector is attached
c. The magnitude of the vector = 60 m
The direction of the vector is southwest = West 45° south
The vector form of the displacement is [tex]\vec{d}[/tex] = 60 × cos(45°)·i + 60 × sin(45°)·j
Which gives;
[tex]\vec{d}[/tex] = 30·√2·i + 30·√2·j
Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)
The drawing of the vector is attached
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A 10.0-kg crate slides along a raised horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second, roughened horizontal surface as shown in the figure. What is the kinetic energy of the crate as it reaches the lower surface?
The kinetic energy of the crate as it reaches the lower surface is 80 J
Kinetic energy is the energy possed by an object in motion. Mathematically, the kinetic energy can be expressed as follow:
KE = ½mv²
With the above formula, the kinetic energy of the crate can be obtained as follow:
Mass (m) = 10 KgVelocity (v) = 4 m/sKinetic energy (KE) =?KE = ½mv²
KE = ½ × 10 × 4²
KE = 5 × 16
KE = 80 J
Therefore, the kinetic energy of the crate is 80 J
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who has the best answer for this
what percentage more water is used to provide us with electricity vs for irrigation
Answer:
electricity because it has more percentage nd energy
Explanation:
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A toaster oven indicates that it operates at 1500 W on a 110 V
circuit. What is the resistance of the oven?
[tex]P=U.I => I=\frac{P}{U}=\frac{1500}{110}=\frac{150}{11}<A>\\I=\frac{U}{R}=> R=\frac{U}{I} = \frac{110}{\frac{150}{11} }=8.06< ohm>[/tex]
The answer is: A. 8.06 ohm
ok done. Thank to me :>
An eagle flying at a constant 120 km/h and has kinetic energy of 2,800 J. What is the mass of the eagle?
Answer:
The mass of the eagle is about 5 Kg
Explanation:
1/2 M= Ke/V^2
120 km/h = 33.3333m/s
1/2 M = 2,800/33.3333^2
1/2 M = 2,800/ 1111.10888889
1/2 M = 2.52000504001
(2) 1/2 M = (2) 2.52000504h001
= M = 5.04001008002
About 5 Kilograms
what is the velocity of 8.01
Answer: The velocity of what?
1. Apply a constant force of 50 N directed to the right of the 50 kg Box. (2 pts)
Hypothesis:?
Conclusion: ?
As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.
Then friction force = 80 Newtons but in the opposite direction.
Friction force = Mu * Normal force exerted by ground = Mu * weight of box
So we find Mu.
Mu = coefficient of friction between box and horizontal surface
= Force of friction / weight = 80 / 50 * 9.81 = 0.163
When an identical box is placed on top, the force of friction is
= Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons
The basic SI unit of length is
Answer:
m
Explanation:
Metres, m, is the SI unit of length
Hope this helped you-have a good day bro cya)
