Explanation:
Thats actually negative acceleration . It means a runner is running with the negative velocity of 1.35 m/s in every second .
the mass of the whole object is equal to the sum of all its parts A. Law of Conservation of Energy B. Law of Conservation of Mass C. Law of Conservation of Weight
HELP PLEASE ILL GIVE YOU BRAINLIEST OR WHATEVER AND LIKE 30 POINTS JUST PLEASE ANSWER ME
Answer:
B. Law of Conservation of Mass
Explanation:
In other words, the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction. If we account for all reactants and products in a chemical reaction, the total mass will be the same at any point in time in any closed system.
HELP ME! PLEASE! SOMEONE! THIS QUESTION IS VERY HARD! THERE ARE 3 ANSWERS! HELP ME PLEASE! MAKE SURE TO EXPALIN
Question:
The teacher told Fatima that all types of energy are kinetic energy, potential energy, or both. Fatima drew this diagram to help her organize this information. She then sorted each type of energy into its correct category: chemical, electrical, electromagnetic, mechanical, nuclear, and thermal.
Three of the types of energy can be either kinetic or potential energy. Identify these three types. For each type, explain why it can be either kinetic or potential.
Answer:
search up the kinetic energy and potential energy etc. then take them and look at the characteristica are they the same? What makes them similar? Why are they different ? How? Then add the chemical nuclear and electrical changes it creates. Now the rest! There you’ve got this! If you need support I’m here! Hope this helped!
Explanation:
Click to review the online content. Then answer the question(s) below, using complete sentences. Scroll down to view additional questions.
Online Content: Site 1
Fitness Log
Name each of the five steps in the PRICE treatment. (Site 1)
Answer:
Protection – Protect the wounded area with a support or something similar.
Rest – Do not continue to exercise while your wound is healing. It may take a few weeks or months to recover. You may need crutches or a sling.
Ice – Wrap a bag of ice or a bag of frozen peas in a towel and apply it to the wounded area for 15–20 minutes every two to three hours.
Compression – Apply elastic compression bandages to the wounded area during the day to reduce swelling.
Elevation – Make sure the injured body part is elevated above the height of your heart as much as possible to help reduce swelling.
Explanation:
Just did it.
Answer:
The five-step process for treating a muscle or joint injury such as an ankle sprain is called "P.R.I.C.E." which is short for Protection, Rest, Ice, Compression, and Elevation).
It can be used even by someone without first-aid training, and should be used immediately when an injury occurs – the earlier, the better – while further medical attention is being sought.
If the athlete experiences too much pain during the process, stop immediately.
"P" is for Protection
Protect the injured person and the area being treated but also protect yourself. If the injury occurs on the sports field, stop the game.
Protect the area being treated with a splint if possible.
If the athlete can move, carefully move them to a safer area using a stretcher or a crutch, but if there is any doubt, do not move the injured athlete.
"R" is for Rest
When a child is injured, the body responds in an effort to defend, localize, protect, and clean up the injured area, a response called inflammation. At the time of injury small blood vessels at the injury site rupture and cause tissue bleeding, which, in turn, can cause bruising and swelling.
Explanation:
There are five warning signs of inflammation:
Pain
Redness;
Tissue hotness;
Swelling, and
Loss of function.
Not every injury exhibits all these signs, but if your child is in pain then it is important to stop exercising immediately to avoid further damage.
"I" is for Ice
As soon as possible after injury, begin applying ice continuously for the first 15-20 minutes to decrease swelling and pain.
An ice bag can be applied directly to the skin (except on the outside of the knee) unless the child has a known cold allergy.
The safest form of icing is to use a plastic bag with ice cubes or crushed ice, or instant cold packs. Do not use “blue ice” or “gel packs”.
While icing will be uncomfortable at first, in 2 to 3 minutes, the skin will go numb and the ice bag will feel more comfortable.
If the child is very uncomfortable with ice bag, then a barrier like a wet towel can be placed between the ice and skin.
The ice bag can be secured in place using an elastic bandage, but the compression should not be too tight.
During the application of the ice, ask the child to wiggle their fingers/toes and monitor tissue around area that is being iced to ensure that they are not experiencing a lost of sensation.
Icing for 15-20 minutes with an ice bag is unlikely to cause nerve damage or frostbite; however, any loss of sensation and any changes in tissue coloration in areas other than the area that is being iced indicate that the application is no longer safe.
Because ice and plastic bags are so important, make sure that someone brings these to every practice and game. Better yet, to be on the safe side, bring your own!
Once a metamorphic rock, always a metamorphic rock. Rocks cannot change from one type of a rock to another.
true or false?
Answer:
False
Explanation:
Metamorphic rock can melt into magma or weathers down (erosion) into sediments.
answer number 19 pls
Answer:
1. Net force = 90 N
2. Acceleration = 9 m/s²
Explanation:
From the question given above, the following data were obtained:
Mass (m) of bag = 10 Kg
Force applied 1 (Fₐ₁) = 45 N
Force applied 2 (Fₐ₂) = 60 N
Force of friction (Fբ) = 15 N
Net force (Fₙ) =?
Acceleration (a) =?
1. Determination of the net force acting on the bag.
We'll begin by calculating the total force applied on the bag.
Force applied 1 (Fₐ₁) = 45 N
Force applied 2 (Fₐ₂) = 60 N
Total force applied = Fₐ₁ + Fₐ₂
Total force applied = 45 + 60
Total force applied = 105 N
Finally, we shall determine the net force acting on the bag. This is illustrated below:
Total force applied = 105 N
Force of friction (Fբ) = 15 N
Net force (Fₙ) =?
Net force (Fₙ) = Total force applied – Force of friction (Fբ)
Net force (Fₙ) = 105 – 15
Net force (Fₙ) = 90 N to the right
2. Determination of the acceleration of the bag.
Mass (m) of bag = 10 Kg
Net force (Fₙ) = 90 N
Acceleration (a) =?
Force (F) = mass (m) × acceleration (a)
90 = 10 × a
Divide both side by 10
a = 90 / 10
a = 9 m/s²
Thus, the acceleration of the bag is 9 m/s² to the right.
what is the direction of the sum of these two vectors?
Answer:
what is the direction of the sum of these two vectors?
Answer:
69.6
Explanation:
Ax= 0
Ay= 63.5
Bx= 101 cos 57= 55.009
By= 101 sin 57= 84.706
Tan-1 = (84.706/55.009)
= 69.6
can somone pls help me??!! i’m very stuck
Answer:
R
Explanation:
Her distance from home and Time increase
the stops when she gets to the library
then her distance from home decreases while her time increases
can I get some help on this
Answer:
sound
Explanation:
but im not exactlly sure so sorry if its wrong.
Answer:
It's Surface wave
Explanation: It travels in a circular motion. The particles travel both parallel and perpendicular in the direction of the wave-
Two football players are attempting to tackle each other. If one football player has a mass of 100 kg and pushes with a force of 100N onto another football player whose mass is 80kg, how much force does the other player push back with?
Answer:
is it 20kg. Two opposing forces pushing onto each other
A projectile was launched horizontally with a velocity of 388 m/s, 2.89 m above the ground. How long did it take the projectile to hit the ground? Select one: 1.34 s 0.308 s 0.768 s 0.0744 s
Answer:
Explanation:
Given
Velocity = 388m/s
Height S = 2.89m
Required
Time
Using the equation of motion
S =ut+1/2gt²
2.89 = 388t+1/2(9.8)t²
2.89 = 388t+4.9t²
Rearrange
4.9t²+388t-2.89 =0
Factorize
t = -388±√388²-4(4.9)(2.89)/2(4.9)
t= -388±√(388²-56.644)/9.8
t = -388±387.93/9.8
t =0.073/9.8
t = 0.00744 seconds
An airplane flies north at 300.0 km/h relative to the air and the wind is blowing south at 15.0 km/h. What is the airplane’s velocity relative to the ground? *
315 km/h North
15 km/h south
285 km/h North
300 km/h South
Answer:
285
Explanation:
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Ahmad is riding his bicycle. He finds that he can accelerate from rest at 0.44 m/s^2 for 5 s to reach a speed of 2.2 m/s. The total mass of Christian and his bicycle is 54 kg. Later, he straps some cargo onto the back of his bicycle. The mass of the cargo is 12 kg.• Calculate the force that Christian can exert on his bicycle before picking up the cargo and exerts the same force on his bicycle.
Answer:
1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo
Explanation:
The given parameters are;
The mass of Christian and his bicycle = 54 kg
The mass of the cargo = 12 kg
1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity
∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo.
Define acceleration. Give its mathematical formula and SI unit. When is acceleration of a body negative? Give two examples of situations in which acceleration of the body is negative.
Answer:
The rate of change of velocity of a moving body is called acceleration. mathematical formula of Acceleration = Final velocity-Initial velocity
________________________
Time taken
SI Unit of Acceleration is m/s2. If the Velocity of a moving body decreases, it's acceleration will be negative. Two examples of situations in which acceleration of the body is negative are ÷
* Applying brakes * When a ball is thrown upward it experiences a negative Accelerationa cyclist accelerates from a initial speed of 0 m/s to a final speed of 8 m/s in 3 second what is his acceleration?
Answer:
3 m/s ^2 rounded to one signifiicant figure.
Explanation:
Hope this helped!
I will give 14 points & make you the brainiest
Terrestrial planets are primarily composed of what two materials?
a. hydrogen and helium
b. silicate rocks and metals
c. frozen water and methane
d. helium and iron
Answer:
B. silicate rocks and metals
Terrestrial planets, such as Earth, are primarily composed of silicate rocks and metals. The correct answer is b. silicate rocks and metals.
Silicate rocks, which are made up of silicon and oxygen, are the dominant building blocks of these planets. They include minerals like quartz, feldspar, and mica. Terrestrial planets contain significant amounts of metals such as iron, nickel, and other heavy elements. These metals are found in the planet's core, providing structural strength and contributing to the planet's magnetic field.
This composition differentiates terrestrial planets from gas giants like Jupiter and Saturn, which are predominantly composed of hydrogen and helium. The combination of silicate rocks and metals forms the solid, rocky surfaces that characterize terrestrial planets.
Therefore, The correct answer is b. silicate rocks and metals.
For more details regarding terrestrial planets, visit:
https://brainly.com/question/7952964
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A child throws a baseball straight up into the air and it takes 4.5 seconds to reach maximum height. What is the total time of flight for the baseball after being thrown and the baseball player catching it? What are the known and unknown variables? And what equation must be used?
Answer:
9 s.
Explanation:
The following data were obtained from the question:
Time (t) taken to reach the maximum height = 4.5 s
Time of flight (T) =?
The time of flight (T) is simply defined as the total time spent by the ball in the air. Mathematically, it defined as twice the time taken (t) to reach the maximum height i.e
T = 2t
With the above formula, we can obtain the time of flight as follow:
Time (t) taken to reach the maximum height = 4.5 s
Time of flight (T) =?
T = 2t
T = 2 × 4.5
T = 9 s
Thus, the time of flight of the ball is 9 s.
An apple and a stone dropped from certain height accelerates:Required to answer. Single choice.Immersive Reader
(1 Point)
equally.
differently.
depending on density.
depending on the position of the sun.
An apple and a stone dropped together from any height accelerate equally.
How do you solving kinematic equations for horizontal projectiles?
An electric heater is rated 1000watts. If a current of 5A pass through the heater, find the value of its resistance.
Answer: R=40 Ω
Explanation: [tex]P=I^{2} R[/tex]
R=1000/ (25)=40
Qui:
3
A fish is swimming forward in the ocean. Which of the following is accurate of the forces
between the fish and the water?
mplete
out of
ove flag
Select one:
O a. The force of the fish's tail on the water is less than the force of the water on the fish's
tail
11
O b. The force of the fish's tail points back while the force of the water points forward
Finis
O c. The force of the fish's tail on the water is greater than the force of the water on the fish's
tail
Time
O d. The forces of both the fish's tail and the water point in the same direction
Check
aus page
Next page
Answer:
O c.the force of the fish tail on the water..........
A 40 kg boy on a skateboard throws a 2 kg ball at 20 m/s to the left. How fast did the boy move to the right on his skateboard?
Find the tension in the two ropes that are holding the 2.9 kg object in place. Rope 1 makes an angle of 51.3° with respect to the vertical. Rope 2 is pulling horizontally.
Answer:
Explanation:
Let the tension in horizontal rope be T₁ and in the other rope be T₂ which is making angle of 51.3⁰
The vertical component of tension T₂ will balance the weight .
= T₂ cos 51.3 = 2.9 x 9.8
T₂ = 2.9 x 9.8 / cos 51.3
= 28.42 / .625
= 45.47 N
The horizontal component of T₂ will balance T₁
T₂ sin 51.3 = T₁
45.47 sin 51.3 = T₁
T₁ = 35.48 N .
What is a transverse wave like?
Answer:
a transverse wave is a moving wave whose oscillations are perpendicular to the direction of the wave or path of propagation
Explanation:
wave oscillate along paths at right angles to the direction of the wave’s advance.
Explanation:
A rock is dropped (from rest) off a bridge over the Merrimack River. The falling rock
accelerates at 10 m/s2 downward. If it takes 2.5 seconds before a splash is heard,
approximately how high is the bridge?
Answer:
31.25 meters or ~31 meters approximately
Explanation:
Let's see which of the 5 variables we are given since this is a constant acceleration problem.
[tex]v_i \ \ \ \ \ \ t \\ v_f \ \ \ \ \ \triangle x \\ a[/tex]We want to find the height of the bridge, aka the vertical displacement of the rock. Let's set the upwards direction to be positive and the downwards direction to be negative.
We are told that the acceleration is 10 m/s² downward, so we have a = -10 m/s².
We are also told that the time it takes the rock to hit the water is 2.5 seconds. Time is the same regardless of the x- or y- direction, so we can say that t = 2.5 seconds.
Now, we aren't told this directly, but we can figure out that the velocity in the y-direction is 0 m/s, since the rock is dropped from rest off the bridge. Therefore, [tex]v_i=0 \frac{m}{s}[/tex].
We want to find the vertical displacement, the height of the bridge, so we can say that [tex]\triangle x= \ ?[/tex]
We have 4 out of 5 variables:
[tex]v_i,\ a, \ t, \ \triangle x[/tex]Look through the constant acceleration equations to see which equation has all 4 of these variables. You should come up with this one (no final velocity):
[tex]x_f=x_i+v_it+\frac{1}{2}at^2[/tex]Subtract [tex]x_i[/tex] from both sides of the equation to get:
[tex]\triangle x=v_it+\frac{1}{2}at^2[/tex]Substitute in our known variables and solve for delta x.
[tex]\triangle x=(0\frac{m}{s})(2.5s) + \frac{1}{2} (-10\frac{m}{s^2})(2.5s)^2[/tex]0 m/s multiplied by 2.5 s is 0, so we have:
[tex]\triangle x =\frac{1}{2} (-10)(2.5)^2[/tex]Evaluate the exponent first and multiply the terms together.
[tex]\triangle x =(-5)(6.25)[/tex] [tex]\triangle x =-31.25[/tex]The vertical displacement is -31.25 meters from the rock's starting position, so we can say that the height of the bridge is 31.25 meters, which is approximately 31 meters tall.
Answer:
31 meters
hope it help you
thanks for free points.........
State the career function of mining engineering
Answer:Mining engineers are responsible for the effective, safe and profitable operation of mining undertakings. They are mining experts and engineers and have a background in geology as well as civil, mechanical and electrical engineering.
Explanation:
The ball rolled 15 meters in 3 seconds. What was the speed of the ball?
Answer:
5 meter / sec.Explanation:
speed = distance / time
= 15 meters / 3 seconds
= 5 meter / sec.
Answer:5/sec
Explanation:
An object has 5m lengh. What does it mean?
Answer:
it means the altitude of the object is 5m high or wide
If it takes Charlie .45 hrs walking at 2.5 km/hr to get to the chocolate factory and it takes Willy Wonka 25 hrs walking at 3.5 km/hr then who lives closer and by how far?
Answer:
6.4
Explanation:
trust
Suppose the mass of the object below was to double. What would happen to the force needed to accelerate the object 10 m/s2?
Answer:
It would also double
Explanation:
F (force) = mass * acceleration; twice the mass means double tue force needed
In a perfectly insulated container of negligible mass, 4.00 × 10−2 kg of steam at 100◦C and atmospheric pressure is added to 0.200 kg of water at 50.0◦C.
A) If no heat is lost to the surroundings, what is the final temperature of the system? B) At the final temperature, how many kilograms are there of steam and how many of liquid water?
Answer:
Following are the solution to this question:
Explanation:
Given value:
[tex]m_s= 4.00 \times 10^{-2} \ kg \\\\L_v=2256 \times 10^{3} \ \frac{J}{kg}\\\\m_w= 0.2 \ kg\\\\\Delta T= 50^{\circ}[/tex]
In point A:
[tex]Q_{Steam}=m_s \ L_v[/tex]
[tex]=0.04 \times 2256 \times 10^{3}\\\\=9.02 \times 10^4 \ J[/tex]
[tex]Q_{water}= m_w \ c_w \ \Delta T\\\\[/tex]
[tex]=0.2 \times 4190 \times 50\\\\=4.19 \times 10^4 \ J[/tex]
[tex]Q_{steam}> Q_{water}[/tex], that's why the final temperature is [tex]= 100^{\circ}[/tex]
In point B:
[tex]\to \Delta m_s L_v=m_w\ c_w \Delta T\\\\\to \Delta m_s \times 2256\times 10^3= 0.2 \times 4190 \times 50\\\\\to \Delta m_s= 1.86 \times 10^{-2} \ kg\\\\\to m_s = 2.14 \times 10^{-2} \ kg\\\\\to liquid \ left = 0.2+ 2.14 \times 10^{-2} = 2.34 \times 10^{-2} \\[/tex]