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Answer:

[tex] \boxed{(i).... 40cm}\\ \boxed{(ii).... 18cm}\\ \boxed{(iii).... 7.5N

}\\ [/tex]

Explanation:

[tex](i).... \\ the \: weight \: of \: the \: rod \: acts \: at \: the \: center \\ which \: is \: the \: half \: of \: 80cm = 40cm \\ (ii).... \\ let \: the \: distance \: be \: \boxed{d}...then \: 22 + d = 40 \\ d = 40 - 22 = 18cm \\ (iii).... \\let \: the \: weight \: of \: the \: metal \: rod \: be \: \boxed{ w} \\ taking \: the \: moment \: about \: the \: string \to \\ 15 \times (22 - 13) = 18w \\ 18w = (9)(15) \\ 2w = 15 \\ w = 7.5N[/tex]

♨Rage♨

Answer:

neutralization reaction

Explanation:

The reaction of an acid with a base is called a neutralization reaction.

es una reacción química que ocurre entre un ácido y una base produciendo sal y agua. ... Se mezcla un ácido fuerte con una base fuerte: Cuando esto sucede, la especie que quedará en disolución será la que esté en mayor cantidad respecto de la otra.

Answer:

I don't know the answer

Answer:

Long question good luck:)..............

Explanation:

Answer: Considering the gravitational field strength being 9.8...

A) 1,120,875J

B) 191.70m/s (2 DP)

Explanation:

explained in pic

You are at the Grand canyon standing at the edge of a ledge 1,857 m high. you have a mass of 61 kg you decide to take a selfie to share with your science teacher when you get home witch causes you to wonder,

a. The gravitational potential energy you have standing on the edge of this cliff is 1,120,875J.

b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, it was felt to be fast as 191.70m/s.

Gravitational potential energy is the potential energy with respect to the gravitational force. To possess gravitational potential energy, the object to be placed in a position in the gravitational field.

According to the question,

Gravitational potential energy = mgh

As given here, h = 1857 m

m = 61

Substituting the given formula,

Gravitational potential energy = 61 × 9.8 × 1875

=1,120,875 J.

Here, As Gravitation potential energy gained, the Kinetic energy lost.

So, Kinetic Energy K.E = 1,120,875 J.

As we know the formula,

K.E = 1/2 mv²

v² =K.E / (1/2 m)

   = 1,120,875 × 2 / 61

v² =36,750

v = [tex]\sqrt{36750}[/tex]

 = 191.702 m/s.

The speed was calculated as 191.702 m/s.

Thus,

a. The gravitational potential energy you have standing on the edge of this cliff is 1,120,875J.

b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, it was felt to be fast as 191.70m/s.

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Answer:

true

Explanation:

Answer:

28.3 m/s

Explanation:

Vi = 45 Km/h = 12.5 m/s

Vf - Vi = at

Vf -12.5 = 0.45(35)

Vf= 28.3 m/s

The answer would be 31 N. This is because Fn = 7.5*9.8*sin(25).

Answer:nfeergnjknrtgrth

Explanation:

Answer:

F = GMm/r^2

MA = GMm/r^2

A = GM/r^2

A = (6.67 * 10^-11)(6.42 * 10^23) / (3.37 * 10^6)^2

A = 3.77 m/s^2

Explanation:

In physics, the triangle symbol (∆) typically represents a change or difference in a particular quantity.



A triangle is often used to denote the difference between two values of a variable. For example, if we have two values of time, t1 and t2, the change or difference in time can be represented as Δt = t2 - t1, where Δt is the triangle symbol indicating the change in time. Similarly, it can be used to represent differences in other physical quantities such as displacement (∆x), velocity (∆v), or temperature (∆T). The triangle symbol (∆) is a shorthand notation commonly used in physics to indicate changes or differences.

In terms of change in displacement, the triangle symbol (∆) represents the difference between two positions or locations. It indicates the change in the object's position from an initial point to a final point.

For example, if an object initially starts at position x1 and then moves to position x2, the change in displacement (∆x) can be calculated as ∆x = x2 - x1. Here, ∆x represents the difference or change in the object's displacement.

The magnitude of ∆x gives the overall distance traveled by the object, and its sign indicates the direction of the displacement. A positive ∆x signifies a displacement in the positive direction (e.g., to the right or upward), while a negative ∆x represents a displacement in the negative direction (e.g., to the left or downward).

Therefore, The triangle symbol (∆) is commonly used in physics equations to denote changes or differences in various quantities, including displacement, velocity, time, and more. It helps us analyze and quantify the differences between two values of a physical quantity.

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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:

Answer:

Yes.

Explanation:

The pressure varies as per the depth of the container

Answer:

a.  -6.99 × 10³⁰ C/m³ b. The layer of air is negatively charged.

Explanation:

With E₁ = electric field at 500 m above the ground = 160 N/C (it is negative since it is directed downwards).

Also, with E₂ = electric field at 800 m above the ground = 120 N/C (it is negative since it is directed downwards).

The total flux, Ψ = ∫E.dA = E₁dAcosθ + E₂dAcosθ

For the 800 m surface E is parallel to dA, that is = 0° and For the 500 m surface E is anti-parallel to dA, that is = 180°  

Ψ = ∫E₁dAcos180° + ∫E₂dAcos0°

= -∫E₁dA + ∫E₂dA

= -E₁∫dA + E₂∫dA

= -E₁4πR² + E₂4πR²

= (-E₁ + E₂)4πR² where R = radius of earth = 6.4 × 10⁶ m

= (-160 N/C + 120 N/C)4π(6.4 × 10⁶ m)²

= - 40 N/C)4π(6.4 × 10⁶ m)²

= -20588.74 × 10¹² C

= -2.058874 × 10¹⁶ C

≅ -2.06 × 10¹⁶ Nm²/C.

The since charge, Q = Ψ/ε₀, the total charge through the area is thus

Q = Ψ/ε₀

= -2.06 × 10¹⁶  Nm²/C ÷ 8.854 × 10⁻¹² F/m

= -0.233 × 10²⁸ C/m²

= -2.33 × 10²⁸ C/m².

So, the charge in the volume = charge net charge of surface × width of volume. So the charge in the volume Q' = QΔh = Q(h₂ - h₁) where h₁ = 500 m and h₂ = 800 m

Q' = Q(h₂ - h₁)

=  -2.33 × 10²⁸ C/m²(800 m - 500 m)

= -2.33 × 10²⁸ C/m²(300 m)

= -699 × 10²⁸ C/m³

= -6.99 × 10³⁰ C/m³

b. Since Q' = -6.99 × 10³⁰ C/m³, the layer of air is negatively charged.

Answer:

I dont really know much but i know that it swallow anything it comes across in space.

Answer:

1) The velocity of the ball return to the thrower's hand is -15 meters per second.

2) The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].

Explanation:

1) Let suppose that ball experiments a free fall, that is an uniform accelerated motion, in which effects from gravity and Earth's rotation can be neglected. The velocity of the ball is represented by the following equations of motion:

Position

[tex]v_{o}\cdot t -\frac{1}{2}\cdot g\cdot t^{2} = 0[/tex]

[tex]t\cdot \left(v_{o}-\frac{1}{2}\cdot g\cdot t \right) = 0[/tex] (1)

Velocity

[tex]v = v_{o}-g\cdot t[/tex] (2)

Where:

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{o}[/tex] - Initial velocity of the ball, measured in meters per second.

[tex]v[/tex] - Final velocity of the ball, measured in meters per second.

From (1), we get the time when the ball returns to the thrower's hand:

[tex]v_{o}-\frac{1}{2}\cdot g\cdot t = 0[/tex]

[tex]t = \frac{2\cdot v_{o}}{g}[/tex]

And then we apply this result in (2):

[tex]v = v_{o}-g\cdot \left(\frac{2\cdot v_{o}}{g} \right)[/tex]

[tex]v = -v_{o}[/tex] (3)

Then, the velocity of the ball return to the thrower's hand is -15 meters per second.

2) The resulting velocity of the boat ([tex]\vec v_{B}[/tex]) is represented by the vectorial sum of the velocity of the boat relative to the river ([tex]\vec v_{B/R}[/tex]) and the velocity of the river ([tex]\vec v_{R}[/tex]), both measured in meters per second, that is:

[tex]\vec v_{B} = \vec v_{R}+\vec {v}_{B/R}[/tex] (4)

If we know that [tex]\vec v_{R} = 6\,\hat{i}\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{B/R} = 18\,\hat{j}\,\left[\frac{m}{s} \right][/tex], then the resulting velocity of the boat is:

[tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex]

The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].

Answer:

stop it get some help

:)

Explanation:

Answer:

My life be like oooh ahhh ooh ah

The correct Answer: would be A B C E

Explanation:

Can i get brainlyest ?

Answer:

The final equilibrium T_{f} = 25.7[°C]

Explanation:

In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.

That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.

The temperature of thermal equilibrium will be T_f.

The heat absorbed by water will be equal to the heat rejected by Iron.

[tex]Q_{iron} = Q_{water}[/tex]

Heat transfer can be found by means of the following equation.

[tex]Q_{iron}=m*C_{piron}*(T_{i}-T_{f})[/tex]

where:

Qiron = Iron heat transfer [kJ]

m = iron mass = 200 [g] = 0.2 [kg]

T_i = Initial temperature of the iron = 300 [°C]

T_f = final temperature [°C]

[tex]Q_{water}=m*C_{pwater}*(T_{f}-T_{iwater})[/tex]

Cp_iron = 437 [J/kg*°C]

Cp_water = 4200 [J/kg*°C]

[tex]0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C][/tex]

Answer:

v₃ = 0.19 [m/s]

Explanation:

To solve this problem we must use the principle of conservation of the linear momentum, which is defined as the momentum before the collision is equal to the momentum after the collision.

ΣPbefore = ΣPafter

[tex]P=m*v[/tex]

P = momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Let's imagine that the toy train moves to the right, this movement is taken as positive.

[tex](m_{1}*v_{1})+(m_{2}*v_{2})= (m_{1}+m_{2})*v_{3}[/tex]

where:

m₁ = mass = 700 [g] = 0.7 [kg]

v₁ = velocity = 0.3 [m/s]

m2 = mass = 400 [g] = 0.4 [kg]

v₂ = 0 (stationary truck, there is no movement)

v₃ = final velocity of the joined engine and truck.

Now replacing:

[tex](0.7*0.3)+(0.4*0)= (0.7+0.4)*v_{3}\\0.21 = 1.1*v_{3}\\v_{3}= 0.19 [m/s][/tex]

Answer:

this is an example of acceleration

Air is cooled and

becomes denser.

Denser air sinks.

The process repeats, producing a (i wasn't sure about this one)

Air is heated and becomes less dense.

Less dense air rises.

hope this helps!

Explanation:

35 maybe hope it's right

Answer:

friction acts upon the ball when it is sliding down the slide and this action creates thermal energy

Answer:

electromagnetic force

Explanation:

Answer:

1.7

2.4

Explanation:

Answer:

1.7

2.4

Explanation:

Answer:Position is an object's location in relation to a reference point. Yes.  Both objects are one unit away (the same distance), but they occupy different locations in space because they are in different directions.

Explanation: Hope this helps!!

Yes it is possible for two objects be at the same distance from a single point but be in different positions.

We have two objects at the same distance from a single point but are in different positions.

We have to determine whether this statement is true or not.

A reference frame can be considered as the coordinate frame with respect to origin of which the physical quantities such as velocity, displacement , acceleration etc. of any other object in different reference frame is measured.

According to the question -

YES, Two objects can have same distance from a single point or a reference point (say origin) and can be at different positions. For example - Two different points with Position vectors -

P(1) = i + j

P(2) = - i + j

are located. Now, distance of point P(1) from origin will be -  [tex]\sqrt{1+1}=\sqrt{2}[/tex] and that of point P(2) from origin will be -  [tex]\sqrt{(-1)^{2} +(1)^{2} } =\sqrt{2}[/tex]

Hence, yes it is possible for two objects be at the same distance from a single point but be in different positions.

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Answer:

the sun is currently a mian sequence star

and will remain so for another 4-5 billion