If 155 g of sugar can be dissolve in 100g of water at 20 degree than how much sugar will be dissolved in 300 g of water at same temperature if someone do I will make you brainly

Answer:

You must label the horizontal axis with the names of the airlines and the vertical axis with the number of flights. The title must clearly state what data the bar chart is showing. With larger numbers, your scale may not go up by one.

Answer:

You must label the horizontal axis with the names of the airlines and the vertical axis with the number of flights. The title must clearly state what data the bar chart is showing. With larger numbers, your scale may not go up by one.

Explanation:

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Answer:

2500 m

Explanation:

Given that,

Each lap is 500 m

Gareth takes 5 laps.

We need to find distance traveled by Gareth. The distance covered by him is given by :

d = 5×500

d = 2500 m

Hence, he will travel 2500 m.

Answer:

2.2 ma

Explanation:

Given :

Length of the rope = L

Mass of the rope = m

Mass of the object pulled by the rope = M

M = 1.5 m

So, L [tex]$\rightarrow$[/tex] m

For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]

∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.3 m

And for remaining 0.7 L =  [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.7 m

By Newtons law of motion,

F - T = ( 0.3 m) a .........(1)

T = ( M + 0.7 m) a

T = ( 1.5 m + 0.7 m) a

T = ( 2.2 m ) a  ..............(2)

So from equation (1) and (2), we have

Tension on the rope

T = 2.2 ma

magnetic domain is a region within a magnetic material in which the magnetization is in a uniform direction. This means that the individual magnetic moments of the atoms are aligned with one another and they point in the same direction.

Answer:

The sun and the stars

Explanation:

I hope this helps!

Answer:

Northstar & North Pole

Explanation:

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Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

Force = 300 N

Find:

Spring constant in N / m

Computation:

Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

Spring constant in N / m = 6,000

Answer:

true

Explanation:

because I tride it

Answer:

The reckage after collision is motionless (E)

Explanation:

The first law of thermodynamics states that energy is neither created nor destroyed but is converted from one form to another.

The kind of collision described in the question above is known as a perfectly inelastic collision, and in this type of collision, the maximum kinetic energy is lost because the objects moving in opposite directions have a resultant momentum that is equal, but in opposite directions hence they cancel each other out.

The calculation is as follows:

m₁v₁ + m₂v₂

where:

m₁ = m₂ = 1000kg

v₁ = 20 m/s

v₂ = -20 m/s ( in the opposite vector direction)

∴ resultant momentum = (1000 × 20) + (1000 × -20)

= 20000 - 20000 = 0

∴ The reckage after collision is motionless

Answer:

The wreckage after collision is moving at the speed 18 m/s to the south.

Explanation:

Answer:

[tex]31.035^{\circ}[/tex]

Explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

y = Displacement in y direction = 0

u = Initial velocity of ball = 7.7 m/s

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Displacement in x direction is given by

[tex]x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}[/tex]

Displacement in y direction is given by

[tex]y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}[/tex]

[tex]\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}[/tex]

The angle at which the ball was thrown is [tex]31.035^{\circ}[/tex].

Answer: Dynamic balance

Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121

Answer:

Bowling ball

Explanation:

Answer:

stop

Explanation:

object stop

Answer:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement. Two or more opposite forces are balanced forces if their effects cancel each other and they do not cause a change in an object's motion.

Answer:

27 ms^-1

Explanation:

by using v= u + at

u = 0 ( because the object id starting from rest)

v= 0 + 1.5 x 18

v = 27 ms^-1

Answer:

F = 2985.125 N

Explanation:

Given that,

The radius of curvature of the roller coaster, r = 8 m

Speed of Micheal, v = 17 m/s

Mass of body, m = 65 kg  We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :

[tex]F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N[/tex]

So, the net force is 2985.125 N.

Answer:

The value 0.51

Explanation:

Answer: 0.33 m/s^2

Explanation:

The acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²

Given the data in the question;

To determine the acceleration of the ship, We the equation from Newton's Second Law of Motion:

[tex]F = m\ *\ a[/tex]

Where F is the force, m is the mass and a is the acceleration

Lets make acceleration ''a'', the subject of the formula

[tex]a = \frac{F}{m}[/tex]

Now, we substitute our given values into the equation

[tex]a = \frac{300000N}{900000kg}[/tex]

We know that, A newton is defined as [tex]1 kg.m/s^2[/tex]

[tex]a = \frac{300000 kg.m/s^2}{900000kg} \\\\a = 0.33m/s^2[/tex]

Therefore, the acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²

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Answer:

 Δt = 5.85 s

Explanation:

For this exercise let's use Faraday's Law

           emf = [tex]- \frac{d \phi}{dt}[/tex] -  d fi / dt

           [tex]\phi[/tex] = B. A

           \phi = B A cos θ

The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.

suppose a linear change of the magnetic field

            emf = - A [tex]\frac{B_f - B_o}{ \Delta t}[/tex]

             Dt = - A  [tex]\frac{B_f - B_o}{emf}[/tex]

the final field before a fault is zero

let's calculate

            Δt = - 0.046 (0- 1.4) / 0.011

            Δt = 5.85 s

Answer:

B = An 8 kg book at a height of 3 m

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

Answer:

[tex]2.4375\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of cannon ball = 39 kg

[tex]m_2[/tex] = Mass of performer = 65 kg

[tex]u_1[/tex] = Initial horizontal component of cannon ball's velocity = 6.5 m/s

[tex]u_2[/tex] = Initial horizontal component of performer's velocity = 0

v = Velocity of combined mass

As the momentum of the system is conserved we have

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{39\times 6.5+0}{39+65}\\\Rightarrow v=2.4375\ \text{m/s}[/tex]

The speed of the performer immediately after catching the cannon ball is [tex]2.4375\ \text{m/s}[/tex].

Answer:

one sec let me think

Explanation:

(a)The average velocity of the ball over the following time intervals will be  [3,4] ft/sec.

(b)The instantaneous velocity at time t = 3 will be32 ft/sec.

(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)The ball will hit the ground at 13.4 sec.

The change of displacement with respect to time is defined as the velocity.  velocity is a vector quantity. it is a time-based component.

The given data in the question will be ,

u is the initial velocity by which ball thrown=128 ft/sec.

V₃ is the instantaneous velocity at time t=3 sec.

V₆ is the instantaneous velocity at time t=6 sec.

t is the time when ball hits the ground=?

(a) Given equation for the displacement

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

Time when velocity is zero will be

[tex]\rm{ t=\frac{128}{32}[/tex]

[tex]\rm{ t=4 sec[/tex]

If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .

Hence the average velocity obtained from the problem will be  [3,4] ft/sec

(b)  

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=3 sec)

v(3) = 128-32×3

v(3) =32 m/sec.

Hence the instantaneous velocity at time t = 3 will be32 ft/sec.

(c)

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=6 sec)

v(6) = 128-32×6

v(6) = -64 m/sec.

Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)

According to Newtons third equation of motion we got

v=u+gt

If the body returens from a certain height at max height its velocity must be zero; ( u=0)

[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]

Hence the ball will hit the ground at 13.4 sec.

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Answer:

B is the Answer

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Mark brainliest

Answer:

The answer is A

Explanation:

This is because f = 1/T,

or frequency equals 1 over period, T

Since the highest T would be the answer with the greatest length and mass does not matter the answer would be c IF they were asking for the highest period.

F is found by 1 over T so the greatest F would be the smallest number, Hence answer A

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Answer: Option b, the final velocity is half of the initial velocity.

Explanation:

Here we will use the conservation of the total momentum of a system.

This means that the total momentum at the beginning must be the same as the final momentum.

Where momentum is:

P = M*v

Initially, we have two cars, both with the same mass M, and only one of them has a velocity v.

Then the initial momentum is:

P = M*v + M*0 = M*v

After the collision, the two cars move together. Then the total mass that is moving is equal to the sum of the masses of the cars, this is 2*M

and we can suppose that the two cars move at a final velocity v'

Then the final momentum is:

P' = (2*M)*v'

Now we use the conservation of momentum, then:

P = P'

M*v = (2*M)*v'

Now we need to solve this for v'

(M*v)/(2*M) = v'

v/2 = v'

This means that the final velocity is half of the initial velocity.

Then the correct option is option b.

Answer: Option b.

Explanation:

We know:

Wavelength = 632.8 nm

Fluence = 4.5*10^20 photon/s

The energy of a single photon of wavelength λ is:

E = (h*c)/λ

where:

h = 6.6*10^(-34) J*s

c = 3*10^8 m/s

And we should rewrite the wave length in meters, so:

λ = 632.8 nm = 632.8*10^(-9) m

replacing these in the energy equation, we get:

E = (6.6*10^(-34)J*s)*(3*10^8 m/s)/(632.8*10^(-9) m) = 3.13*10^(-19) J

So each one of the  4.5x10^20 photon that flow each second have this energy, then the power is:

P = (3.13*10^(-19) J)*(4.5*10^20 /s) = 140.85 J/s

and 1 W = 140.85 J/S

Then the power is:

P = 140.85 W

Then the correct answer is the option b, where the units are a little bit different than mine because I used really simplified values for c and h.

During the "waxing" phases, the two weeks immediately following the New Moon.

New Moon ==> waxing crescent ==> First Quarter ==> waxing gibbous ==> Full Moon

The phase of the moon decides the moon's visibility. The sequence of phases does the moon's visibility increases is; the waxing crescent, first quarter, waxing gibbous, full Moon.

The moon changes shape every day. This is due to the fact that the celestial body has no light of its own and can only reflect sunlight.

Only the side of the moon facing the sun can reflect this light and seem bright. The opposite side appears black. this is a full moon.

We can only see the black section when it lies between the sun and the earth when a new moon occurs. We witness intermediate phases like a half-moon and crescent in between these two extremes.

Following are the sequence of phases does the moon's visibility increase is;

1. Waxing crescent

2. First Quarter

3. Waxing gibbous

4. Full Moon

The phase of the moon decides the moon's visibility. Hence the visibility changes with the change in the phase of the moon.

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