Answer:
To Avoid accidents and injuries
Explanation:
Safety measures are vital dancing as this may prevent bad things like accidents and injuries from happening.
When dancing,the safety of dancers is a must to be observed and considered.
Dancers are also most likely to exert their full potential if they know that they are safe no matter what happen....
HAVE A GOOD DAY....
HELP PLEASE
a person standing on the roof of a building throws a rubber ball down with a velocity of 8m/s
what is the acceleration(magnitude and direction) of the ball? does the ball slow down while falling? after 0.25 seconds how far has the ball fallen?
Answer:
2.30625m
Explanation:
The acceleration of the ball is the acceleration of gravity g.
By definition
g = constant = + 9.8 m?s^2 and is directed towards the centre of the earth
The ball does not slow down while falling
The position function is:
P(t) = 1/2 a t^2 + Vo t
In this specific case: P(t) = 4.9 t^2 + 8t
P(0.25s) = 4.9 (0.25)^2 + 8*0.25 = 2.30625m
WILL MARK BRAINLIEST FOR CORRECT ANSWERS
A rocket will accelerate...
Answer:
True
False
False
Explanation:
The first answer is inferred from Newton's third law of motion. As the rocket burns fuel and ejects the products downwards, a thrust is created, which acts downwards; pushing the rocket upwards.
The second answer is gotten from the same law. Pushing against the atmosphere (action) will create a reaction (this reaction is from the atmosphere; action and reaction are equal and opposite) which acts downwards. Hence the rocket doesn't move up.
Third. When the rocket is in space (vacuum), it moves at a uniform velocity since no force acts on it. This is gotten from Newton first law of motion. Here, the rocket moves into the orbit, fuelled by the force of it's thrust (when it was on the ground).
Bob creates an instrument that is able to play C4 (261.63 Hz). He does some analysis with the sound equipment and it shows that when he plays it, he also gets the frequencies 523.26 Hz, 784.89 Hz, and 1308.15 Hz. What could be true about the instrument? It’s a stringed instrument, it’s a closed pipe, or it’s an open pipe?
Answer:
It could be a stringed instrument or an open pipe.
Explanation:
Let v be the speed of sound, y be wavelength and f be frequency.
v = yf
f= v/y; v is constant.
In a stringed instrument, the fundamental frequency note is heard when the length of the string, l = y/2; y= 2l
f′= v/2l
The second harmonic is heard when l= y
f"= v/y
...
f'''= 3v/2l
We can infer that f"= 2f'
f'''= 3f'
This is similar to the values in the question as;
523.26 =2(261.63) and so on.
Same thing happens with open pipes.
Refer to the image shown to answer the question.
If a light ray strikes the shiny surface at 70 degrees, what is the measurement of the reflected ray?
170 degrees
110 degrees
70 degrees
10 degrees
Answer:
the reflected ray will be 110 degree
What are some uses of thermal energy in everyday life?
Choose all that apply.
moving objects
cooking food
heating water
making light bulbs work
Answer:
b) cooking food
c) heating water
Explanation:
cooking and heating use thermal energy to transfer heat from the source to another object, in the examples provided; the food and the water. on the other hand, moving objects uses kinetic energy and lightbulbs utilize electrical energy.
Answer:
cooking food
heating water
i think making a light bulb work
Explanation:
According to the HR diagram below, blue giants are in what temperature range?
A. 2,500 k - 5,000 k
B. 5,000 k - 10,000 k
C. 10,000 k - 20,000 k
D. 20,000 k - 40,000 k
Answer:
20 000 - 40 000 K from the graph
Explanation:
Look straight down from blue giants and read the temps
55. A projectile of mass 2.0 kg is fired in the air at an angle of 40.0° to the horizon at a speed of 50.0 m/s. At
the highest point in its flight, the projectile breaks into three parts of mass 1.0 kg, 0.7 kg, and 0.3 kg. The 1.0-
kg part falls straight down after breakup with an initial speed of 10.0 m/s, the 0.7-kg part moves in the original
forward direction, and the 0.3-kg part goes straight up.
At peak after explosion
V3,f = (V3,1))
At peak before explosion
V₁
=
(V₁.x)x
m3 = 0.3 kg
m₂ = 0.7 kg
V2,1 = (v₂,1)
m₁ = 1.0 kg
V₁ = (50.0 m/s)v₁
₁,f= -(10.0 m/s)
40°
Launch
a. Find the speeds of the 0.3-kg and 0.7-kg pieces immediately after the break-up.
b. How high from the break-up point does the 0.3-kg piece go before coming to rest?
c. Where does the 0.7-kg piece land relative to where it was fired from?
The total moment in an islated system remains constant through time. a) V₂ = 109.43 m/s and V₃ = 33.33 m/s. b) h = 56.67 m. c) d = 480 m.
What is the law of conservation of momentum?First let us remember that momentum or lineal momentum is a motion quantity. In physics it is the fundamental quantity that characterizes the motion of any object.
The momentum is a vectorial quantity that can be calculated as the product of the object mass by its lineal velocity ⇒ mv
The total lineal moment of a system constituted by a group of objects is the sum of the vectorial moments of each of the objects.
In an isolated system -the one that does not interact with the exterior environment-, the moment remains constant through time. This is the law of conservation of momentum. The initial moment is equal to the final moment ⇒ Pinitial = Pfinal
Objects involved in crushes, explosions, collitions, and others, are considered to be isolated systems.
.......................
So, in the exposed example we know that,
Projectile mass ⇒2kg.Velocity = 50 m/s.Angle⇒ 40°.Smaller masses ⇒ 1 kg with 10m/s speed, 0.7 kg, and 0.3 kg.a) We need to calculate the speed of the 0.7 and 0.3 kg objects.
So, 0.7 kg object moves in the original forward direction, meaning that we can use the following formula to calculate its velocity,
m₀v₀ = m₂v₂
Where
m₀ = projectile mass = 2kgv₀ = horizontal component of projectile velocitym₂ = object mass = 0.7 kgv₂ = object velocity = ??So first, we need to get the horizontal component of projectile velocity, V₀.
V₀ = V cosθ
V₀ = 50 cos40º
V₀ = 38.3 m/s
Now we can calculate the velocity of the 0.7 kg object.
m₀v₀ = m₂v₂
2 kg x 38.3 m/s = 0.7 kg x v₂
v₂ = (2 x 38.3) / 0.7
v₂ = 109.43 m/s
Now we need to calculate the velocity of the 0.3 kg object. Since this part goes exactly in the opposite direction of the 10 kg object, their addition will equal zero. So we can use the following equation to calculate the velocity of the 0.3 kg object.
m₁v₁ + m₃v₃ = 0
Where
m₁ = 1 kg object mass v₁ = 1 kg object velocity = 10m/sm₃ = 0.3 kg object mass v₃ = object velocity = ??m₁v₁ + m₃v₃ = 0
(1 kg x 10m/s) + (0.3 kg x v₃ ) = 0
(0.3 kg x v₃ ) = - (1 kg x 10m/s)
v₃ = - (1 kg x 10m/s) / 0.3
v₃ = - 33.33 m/s
The - sign is indicating the opposite direction concerning the 1 kg object.
ANSWER:
The speed of the 0.7 kg object is 109.43 m/sThe speed of the 0.3 kg objects is 33.33 m/s.........................
b) We need to get the high from the break-up point where the 0.3-kg piece go before rest
At rest,
1/2 m₃v₃² = m₃gh
Where,
m₃ = 0.3 kg object mass v₃ = object velocity = 33.33m/sg = gravity = 9.8 m/s²h = high = ??1/2 m₃v₃² = m₃gh
h = 1/2 (v₃²) /g
h = 1/2 (33.33²) / 9.8
h = 56.67 m
ANSWER: The high from the break-up point where the 0.3-kg piece go before coming to rest is 56.67 m.
........................
c) Finally, we need to get where does the 0.7-kg piece land
ANSWER: The distance at which the 0.7 piece lands, concerning the point from which it was fired is 480 m.
You can learn more about law of conservation of momentum at
https://brainly.com/question/17140635
https://brainly.com/question/1113396
#SPJ1