Answer:
mas lalo kang tinatamaran magaral
A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary on a surface with a coefficient of kinetic friction of 0.15 when hit, how far does it move after the bullet emerges?
The distance traveled by the wood after the bullet emerges is 0.16 m.
The given parameters;
mass of the bullet, m = 23 g = 0.023 gspeed of the bullet, u = 230 m/smass of the wood, m = 2 kgfinal speed of the bullet, v = 170 m/scoefficient of friction, μ = 0.15The final velocity of the wood after the bullet hits is calculated as follows;
[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s[/tex]
The acceleration of the wood is calculated as follows;
[tex]\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2[/tex]
The distance traveled by the wood after the bullet emerges is calculated as follows;
[tex]v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m[/tex]
Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.
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A 3000 kg car stops at a red light, and is rear-ended by a 5000 kg truck traveling at 20m/s. In the collision, the two cars stick together. What is the final speed of the two cars just after the collision in m/s (Numeric Answers only)
Explanation:
this is actually not as simple as it sounds here.
quite some energy is lost in the deformation of the bodies of car and truck, and it also needs more energy to get a standing object going than to accelerate an already moving object.
but assuming the simple described circumstances, then the energy and impulse of the moving truck of 5000 kg is transferred to a new combined system of car and truck of now 5000 + 3000 = 8000 kg.
so, the 20m/s inertia energy of the truck is now distributed to the truck/car combination.
since the same energy has to move now more mass, it is clear that the combined speed will be lower.
20×5000 = x×8000
20×5 = x×8
x = 100/8 = 12.5 m/s
that is the resulting speed of the combined truck/car object.
A person is driving a car in a roundabout at 20 m/s so that its centripetal acceleration is 5 m/s2 . How much time it takes to complete 1 circle around the roundabout?
Answer:
Ac = v²/r
where, Ac is centripetal acceleration
v is velocity
r is radius
5 = (20)²/r
r = 400/5
r = 80m
from,
w = 2πf
where, w = angular velocity and f frequency
w = v/r
f = n/T
then,
v = 2πrn ,n = 1 to complete 1 cycle
T
T = 2πr/v
T = 2π*80/20
T = 25.14seconds.
The car would take 25.14s to complete 1 cycle of the roundabout.
Pls help need asap thank u!
Answer:
c
Explanation:
10. Unless a light ray comes into contact with a surface or enters a different material, it
travels in a
Answer:
Straight Path
Ex 3) A tennis player lobs the tennis ball up in the air during a serve at a rate 3 m/s.
How long will it take the tennis ball to reach its maximum height?
A__________ shows how velocity changes over time
HELPPP
Which of the following statements about friction is FALSE .
1 ) Friction involves objects that are in physical contact .
2) Friction is less if surfaces are very smooth .
3 ) Friction is always in the direction of the motion .
4) Friction is caused by the uneven surfaces of touching objects
define potential difference as used in electricity
Answer:
Explanation:
Testing answer please do not delete
the diffrence in potential betwen two points that represents the work involved or the energy relesed in the transfe of a unit quantity of electicity from one point to the other
ALREADY SOLVED Thank you for looking.
Answer:
yes
Explanation:
A spring of spring constant k = 200 N m−1 is slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. Calculate the work done by the extending force. 60
Answer:
31
Explanation:
No need
Un avión vuela horizontalmente con una velocidad de 800 km/h y deja caer
un proyectil desde una altura de 500 respecto al suelo. a) ¿Cuánto tiempo
transcurre antes de que el proyectil se impacte en el suelo?; b) ¿Qué distancia
horizontal recorre el proyectil después de iniciar su caída?
The projectiles launch allows to find the results for the questions of the launch of the bomb are:
a) The fall time is: t = 10.1 s
b) The distance traveled is: x = 2.24 10³ m
Projectile launching is an application of kinematics where the acceleration on the x-axis is zero and the acceleration on the y-axis is gravity acceleration.
a) Let's find the time until we reach the ground.
The initial vertical velocity is zero, the initial height is I = 500 m, when reaching the ground its height y = 0.
[tex]y = y_o + v_o_y - \frac{1}{2} g t^2[/tex]
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{\frac{2y_o}{g}[/tex]
t = [tex]\sqrt{\frac{2 \ 500}{9.8} }[/tex]
t = 10.1 s
b) We look for the horizontal distance traveled.
Let's reduce to the international measurement system (SI).
v = 800 km / h ([tex]\frac{1000m}{1km}[/tex]) ([tex]\frac{1h}{3600s}[/tex]) = 222.22 m / s
x = vₓ t
x = 22.22 10.1
x = 2.24 10³ m
In conclusion, using the projectile launch relationships, we can find the results for the questions that are:
a) The fall time is: t = 10.1 s
b) The distance traveled is: x = 2.24 10³ m
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When converted to a household measurement, 9 kilograms is approximately equal to a
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
What process makes this pattern on the ocean floor?
When converted to a household measurement, 9 kilograms is approximately equal to a
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
4. What is the density of a block with a mass of 36 g and a volume of 9 cm?
O A 45 g/cm3
O B.27 g/cm
O 0.4 g/cm
O D. 0.25 g/cm
Answer:
0.4 g/cm
Explanation:
density (g cm ³) = mass (g)
÷
volume (cm³)
Answer:
C. 4 g/cm
Explanation:
Use the formula:
density = mass ÷ volume
mass = 36volume = 9Sub in the values:
density = 36 ÷ 9 = 4 g/cm
Answer = 4 g/cm
*Sorry for the bad quality picture!*
A frictionless pendulum with a mass of 0.4 kg and a length of 2.1 m starts at point A, at an angle 0 of 60°. As it swings downward, it passes through point B, which is 30 degrees from equilibrium. What is the kinetic energy of the pendulum at point B?
A) 3.9 J
B) 3.0 J
C) 1.1 J
D) 4.1 J
The conservation of mechanical energy allows finding the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
The conservation of mechanical energy is a theorem of greater importance in physics and ordinary life, it states that if there is no friction force the total mechanistic energy remains constant at all points.
Mechanical energy is the sum of kinetic energy plus all potential energies. In the attachment we see a diagram of the pendulum's movement at the two points of interest.
They indicate that the pendulum is released from an initial angle of θ₁ = 60º, let's find the mechanical energy at that point.
Em₀ = U = m g h
Where the height is measured from the lowest point of the movement.
h = L - L cos tea1 = L (1 cos tea1)
The second point of interest occurs for θ₂ = 30º.
At this point part of the energy is indica and part gravitational potential.
[tex]Em_f[/tex] = K + U₂
[tex]Em_f[/tex] = ½ m v² + m g h ’
There is no friction in the system, therefore mechanical energy is conserved.
Em₀ = Em₀_f
mg L (1 - cos θ₁) = ½ m v² + m g L (1 - cos θ₂)
v² = 2g L (cos θ₂ - cos θ₁)
Let's calculate.
v² = 2 9.8 2.1 (cos 30 - cos 60)
v² = 41.16 0.366
v = 3.88 m / s
In conclusion using the conservation of mechanical energy we can find the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
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Answer:
3.0 J
Explanation:
Just took the test
The volume of an ideal gas is increased from 0.6 m3 to 2.4 m3 while maintaining a constant pressure of 1000 Pa (1000 N/m2). Determine, in J, the amount of work done by the gas in this expansion.
The amount of work done by the gas in the given expansion is 1800 J.
The given parameters;
initial volume of the ideal gas, V₁ = 0.6 m³final volume of the ideal gas, V₂ = 2.4 m³constant pressure of the gas, P = 1000 PaThe amount of work done by the gas in the given expansion is calculated as follows;
W = PΔV
where;
ΔV is the change in volume of the gasSubstitute the given parameters and solve for the work done;
W = 1000(2.4 - 0.6
W = 1800 J
Thus, the amount of work done by the gas in the given expansion is 1800 J.
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10 basic rules of badminton?
Explanation:
Rules
A match consists of the best of three games of 21 points.
The player/pair winning a rally adds a point to its score.
At 20-all, the player/pair which first gains a 2-point lead wins that game.
At 29-all, the side scoring the 30th point wins that game.
The player/pair winning a game serves first in the next game.
A badminton match can be played by two opposing players (singles) or four opposing players (doubles).
A competitive match must be played indoors utilising the official court dimensions.
A point is scored when the shuttlecock lands inside the opponent's court or if a returned shuttlecock hits the net or lands outside of the court the player will lose the point.
At the start of the rally, the server and receiver stand in diagonally opposite service courts.
A legal serve must be hit diagonally over the net and across the court.
A badminton serve must be hit underarm and below the server's waist height with the racquet shaft pointing downwards, the shuttlecock is not allowed to bounce. After a point is won, the players will move to the opposite serving stations for the next point.
The rules do not allow second serves.
During a point a player can return the shuttlecock from inside and outside of the court.
A player is not able to touch the net with any part of their body or racket.
A player must not deliberately distract their opponent.
A player is not able to hit the shuttlecock twice.
A 'let' may be called by the referee if an unforeseen or accidental issue arises.
A game must include two rest periods. These are a 90-second rest after the first game and a 5-minute rest after the second game.
Ex 1) A Major League baseball pitcher throws a baseball straight up into the air and the ball travels for 4
seconds reaching a height of 52 m before returning to the pitchers hand exactly at the height that he released
it. With what velocity does the baseball strike the pitchers hand on the way back down?
Answer:
52 / 4 = 13 m/s
Explanation:
Ex 3) A tennis player lobs the tennis ball up in the air during a serve at a rate 3 m/s.
How long will it take the tennis ball to reach its maximum height?
Explanation:
s=(v^2-u^2)/2g
=(0-3^2)/2*10
= 0.45m
What element is chemically similar to carbon
nonmental
that should be the answer
Computes average speed when a person walk for 1 minute at speed of 1.5 m/s and another 1 minute at speed of 3.5 m/s along a straight road
Answer:
2.5 m/s
Explanation:
(60*1.5+60*3.5)/120=2.5
or
(1.5+3.5)/2=2.5
Suppose a ball is dropped from a building. What is the change of velocity between the first and fifth seconds of its flight
Answer: 39.2m/s
Explanation:
Due to the free falling body, the acceleration of the ball is 9.8 m/s^2, and its initial velocity is 0 m/s.
V(t) = g*t
The change of velocity from t=1 to t=5:
Delta V = 9.8*(5-1) = 39.2 m/s
There are 6 foundation of sports and which one you think is the most important?
I just need the points
Explanation:
But just pick any and say something like "it stans out to me most/more" or "it sounds/looks more intristing to me"
1) A bowling has a mass of 5 kg and a speed of 8 m/s. What is its momentum?
2) If the bowling ball started from a momentum of O and reached a momentum of 40
kg*m/s in 2 seconds. What is the impulse impacted on the bowling ball?
3) What is the force of impact for the 2 seconds?
Based on your observations, what can you say about your prediction in Parts A and B above concerning the potential and kinetic energy?
Answer:
Explanation:
My perdiction was correct. I predicted in Part A that the Kinetic Energy would increase. In Part B I predicted that the potential energy would decrease due to the kinetic energy increasing.
A machine whose efficiency is 75% is used to lift a load of 100m.calculate the effort put into the machine if it has a velocity ratio of 4.
Explanation:
M.A = load / Effort
efficiency = M.A/V.R X 100
75 = M.A / 4 X 100
75 = 25 X M.A
M.A = 75/25 = 3
M.A = load / effort
3 = 100/E
E = 100/3 = 33.333
A machine whose efficiency is 75% is used to lift a load of 100m. the effort needed to put into the machine is 33.33 meters if the velocity ratio is 4.
What is the mechanical advantage?Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.
Mechanical advantage = output force(load) /input force (effort)
Simple machines like static pulleys, as well as a movable pulleys, provide a great amount of mechanical advantage.
A machine whose efficiency is 75% is used to lift a load of 100m.calc if it has a velocity ratio of 4.
As we know that the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.
mechanical advantage = velocity ratio ×Efficiency
=4 × 0.75
= 3
The mechanical advantage is 3 , which means the ratio of load to effort is 3
load / effort = 3
100 m/effort =3
effort = 33.33 m
Thus, the effort put into the machine comes out to be 33.33 m.
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What is the value of the force measured by the spring balance?
Someone help this is a test and i need to finish this quick i will give brainliest of it lets me
Sorry, I don't know but I think the correct answer is the first option.