Use words from the box to complete the sentences below.
-direction
-energy
-mass
-size
Vectors have size and ___
Scalars have only ____

Answer:

Explanation:

speed = distance / time

          = 15 meters / 3 seconds

          = 5 meter / sec.

Answer:5/sec

Explanation:

b=36•3

momentum=mass×velocity

The total momentum of the given system is equal to zero. Therefore, option (C) is correct.

The linear momentum can be described as the product of the mass times the velocity of that object. Conservation of momentum is a property of an object as the total amount of momentum stays the same.

According to the law of conservation of momentum, the sum of the momentum before and after the collision of the objects must be equal.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂v₂

where u₁ and u₂ are initial speed while v₁ & v₂ is final speed and m₁ and m₂ is the mass of the collided objects.

The first ball, m = 8.52 g and v = 2.13 m/s

The momentum of the first ball = 18.15 g.m/s

The second ball is moving in the opposite direction w.r.t. first ball,

The second ball, m = 8.52 g and v = - 2.13 m/s

The momentum of the first ball = - 18.15 g.m/s

The total momentum  of the system = 18.15 + (-18.15) = 0

Therefore, the total momentum of the given system is zero.

Learn more about the law of conservation of linear momentum, here:

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Answer: R=40 Ω

Explanation: [tex]P=I^{2} R[/tex]

R=1000/ (25)=40

Answer:

B. helium, xenon, neon

Explanation:

Answer:

i guss its A

Explanation:

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:

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

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 ÷

Answer:

The correct option is;

c. 22.6

Explanation:

The given parameters are;

The hypotenuse of the vector = 32

The angle of the vector = 45°

Therefore, the vector component in the y-axis is given as follows;

[tex]v_y = v \times sin(\theta)[/tex]

Substituting the values from the question gives;

[tex]v_y = 32 \times sin(45^{\circ}) \approx 22.6[/tex]

The vector component in the y-axis, [tex]v_y[/tex], is approximately 22.6.

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

Answer:

False

Explanation:

Metamorphic rock can melt into magma or weathers down (erosion) into sediments.

Answer:

285

Explanation:

read and alot jbgdxtffhgdae

Answer:

d

Explanation:

42000 kilometers

Explanation:

distance is velocity × time

28000 ×1.5

42000

remember to match units

thats why i wrote 1.5 hours instead of 90minutes

Answer:

O c.the force of the fish tail on the water..........

Answer:

Option (a) is correct.

Explanation:

Let the magnitude of the positive charge = [tex]q_1[/tex] and the magnitude of the negative charge = [tex]q_2[/tex]

Earlier, the distance between both the charge, d = 1m

By using Coulomb's law, the magnitude of the force between two charge

[tex]F=k\frac {q_1 q_2}{d^2}[/tex], where k is a constant.

So, the magnitude of the force in the initial configuration,

[tex]F_i = k\frac {q_1 q_2}{1^2}= k q_1q_2\cdots(i)[/tex]

The nature of the force is attractive, as both the charges are opposite in nature.

On replacing the positive charge from 1m left to 1 m right side of the negative charge, the distance between the charges remains the same, i.e d=1m.

Moreover, the magnitude, and nature of each charge, [tex]q_1[/tex] as well as [tex]q_2[/tex], are remain the same.

So, the magnitude of the force in the final configuration,

[tex]F_f = k\frac {q_1 q_2}{1^2}= k q_1q_2[/tex]

From equation (i), [tex]F_f=F_i[/tex]

The nature of the force is attractive, as both the charges are opposite in nature.

So, the electrostatic force remains attractive, and the magnitude does not change.

Hence, option (a) is correct.

When the electrostatic force lies between the charges so here the force should remain attractive, also the magnitude should remain the same.

Here we assume the magnitude of the positive charge should be q1 and the magnitude of the negative charge should be q2

Also, the distance between the charge should be d = 1 m

So here we use the columb law

F = kq1q2/d2

here k should be constant

Here the nature of the force should be attractive since the both the charges should be opposite.

Hence, the first option is correct.

Learn more about force here: https://brainly.com/question/3398162

Answer:

F = 800 [N]

Explanation:

To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.

We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.

ΣPbefore = ΣPafter

[tex](m_{1}*v_{1}) - F*t = (m_{1}*v_{2})[/tex]

where:

m₁ = mass of the hammer = 0.15 [m/s]

v₁ = velocity of the hammer = 8 [m/s]

F = force [N] (units of Newtons)

t = time = 0.0015 [s]

v₂ = velocity of the hammer after the impact = 0

[tex](0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N][/tex]

Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.

Answer:

It would also double

Explanation:

F (force) = mass * acceleration; twice the mass means double tue force needed

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]

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.

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.

Answer:

resultant force=16N→(TOWARDS THE DIRECTION OF 8N)

Explanation:

resultant force=8N+13N-5N

                        =16N→(TOWARDS THE DIRECTION OF 8N)

Answer:

is it 20kg. Two opposing forces pushing onto each other

Answer:

Following are the solution to this question:

Explanation:

That light takes a very long time to hit the planet, and the object is far off the earth. The light of such an item near to the planet takes less time to enter it. The star is 2,5 million light-years from the Planet on the far side of the Andromeda Galaxy. But on the other hand, the moon is 15 crore miles from the earth, so sunlight is quickly reached on the ground as the other thing.  

That milky way away from the earth is 66,500 light-years far, that distance between Earth and Orion nebula is 1,344 light-years, with such a distance of 4,367 light-years. The earth is 5.2261 trillion km apart from Pluto.

Answer:

    m v ´- MV = (m + M) v ’

Explanation:

If the astronaut initially has a speed v the satellite has a speed V, we can define a system that is formed by the two bodies, therefore the forces during the collision are internal, so the momentum is conserved

initial instant. Before the crash

          p₀ = m v - MV

final instatne. After the crash, we have two cases

1) inelastic shock

          m_f = (m + M) v '

          p₀ = p_f

         m v ´- MV = (m + M) v ’

2) elastic collision with the astronaut's velocity, zero in this case the moment remains

        m v - MV = mv ’+ Mv’

also the kinetic energy is conserved

       mv'2 + M V2 = mv'2 + m v'2

with these two equations we can find the speed of the cars

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.

Question:

A cork floats on the surface of an incompressible liquid in a container exposed to atmospheric pressure. The container is then sealed and the air above the liquid is evacuated. The cork:

A. sinks slightly  

B. rises slightly  

C. floats at the same height  

D. bobs up and down about its old position

Answer:

The correct answer is C)  floats at the same height  

Explanation:

The liquid is incompressible because its density very high and leaves no room for further compaction whether or not there is atmospheric pressure. So when you put a cork on the liquid, pressure or no pressure, there is no displacement hence it floats on the same height regardless of the absence of air.

Cheers!

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!