which statement is true????

Answer:

Explanation:

The average speed of the ball can be found by using the formula

[tex]v = \frac{d}{t} \\ [/tex]

d is the distance

t is the time taken

From the question we have

[tex]v = \frac{50}{5} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

[tex]F=8.64\times 10^{-9}\ N[/tex]

Explanation:

Given that,

Mass of student 1, m₁ = 120 kg

Mass of student 2, m₂ = 108 kg

Let they are at a distance of 10 m.

We need to find the gravitational pull they have on each other. The gravitational force acting between two objects is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{120\times 108}{(10)^2}\\\\=8.64\times 10^{-9}\ N[/tex]

So, the gravitational pull between students is [tex]8.64\times 10^{-9}\ N[/tex].

Answer:

Entonces la velocidad es 2 [tex]\frac{m}{s}[/tex] hacia la derecha, mientras que la rapidez es 2 [tex]\frac{m}{s}[/tex].

Explanation:

La rapidez es una magnitud escalar que relaciona la distancia recorrida con el tiempo mientras que la velocidad es una magnitud vectorial que relaciona el cambio de posición (o desplazamiento) con el tiempo.

Es decir, la rapidez es una magnitud física que expresa el valor numérico y la unidad de una distancia en relación con el tiempo.

Matemáticamente,  es expresada como el cociente entre el camino recorrido y el tiempo transcurrido:

[tex]rapidez=\frac{distancia}{tiempo}[/tex]

Por otro lado, la velocidad expresa cómo se está moviendo un objeto en cada momento, informando la dirección, el sentido y la rapidez del movimiento (la velocidad se refiere al intervalo de tiempo que le toma a un objeto desplazarse hacia una dirección determinada y, al involucrar la dirección o sentido del movimiento, es una magnitud vectorial).

Matemáticamente, la velocidad es calculada como el cociente entre el desplazamiento que realizó un cuerpo  y el tiempo total  que le llevó realizarlo.

[tex]velocidad=\frac{desplazamiento}{tiempo}[/tex]

En este caso, sabes que:

Reemplazando obtenes que la aceleración es:

[tex]rapidez=\frac{10m}{5 s}[/tex]

rapidez= 2 [tex]\frac{m}{s}[/tex]

y la velocidad es:

[tex]velocidad=\frac{10m}{5s}[/tex]

velocidad= 2 [tex]\frac{m}{s}[/tex] hacia la derecha.

Entonces la velocidad es 2 [tex]\frac{m}{s}[/tex] hacia la derecha, mientras que la rapidez es 2 [tex]\frac{m}{s}[/tex].

Answer:

12 A

Explanation:

Given,

Power of electric kettle=2.2 kW=2200W (since 1kW=1000W)

Voltage=180V

Current=?

Power = Voltage*Current

Current = Power ÷ Voltage

=2200W ÷ 180V

=12.22 Ampere

=12 A

The amount of current moving through the electric kettle with the given power and voltage is 12A.

Given the data in the question;

Using the the power law equation:

Electrical power is the product of voltage and current:

[tex]P = V * I[/tex]

Where V is voltage and I is current

We substitute our given values into the equation

[tex]2200W = 180V * I\\\\I = \frac{2200W}{180V} \\\\I = 12.22A\\\\I =12A[/tex]

Therefore, the amount of current moving through the electric kettle with the given power and voltage is 12A.

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Answer:

0.2

Explanation:

3/15

Answer:

Kinetic energy is the energy obtained from movement while potential energy is the energy in something when it is at a height or has been manipulated in some way; it has work done to it.

Explanation:

Kinetic energy is, like I said, energy gained through movement or action. When an object stops moving, it converts into a different type of energy. Energy cannot be destroyed or created. An object with potential energy has a reference. A reference determines how high it is. The higher and heavier an object is, the more potential energy it has.

Hope this helps!

Answer:

False

Explanation:

The second you let go its gonna release kinetic energy that's why it's potential

The time taken for the heavy car to reach the bottom of the frictionless slope is 0.56 s.

The height of the incline is calculated as follows;

sin θ = h/L

h = L sinθ

where;

h = 17.5 x sin(5.09)

h = 1.55 m

t = √(2h/g)

where;

t = √(2 x 1.55 /9.8)

t = 0.56 s

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Answer:

T=  6.34

Explanation:

This is the answer. This answer also works on acellus

You push against a wall, and the wall applies a force to your  hands. (A)

Answer:

The largest surface area window that can be used is [tex]A=0.01\ m^2[/tex]

Explanation:

Pressure

The pressure is defined as force per unit area. The SI unit for pressure is the Pascal (Pa), defined as Newton per square meter.

If a force F acts on a surface area A, the pressure is calculated as:

[tex]\displaystyle P=\frac{F}{A}[/tex]

The pressure at a depth of 90 m is P=1 Mpa= 1,000,000 Pa and the submarine's window can only take a force of F=10 kN=10,000 N.

We need to calculate the largest surface area of the window. We can solve the equation for A:

[tex]\displaystyle A=\frac{F}{P}[/tex]

Substituting:

[tex]\displaystyle A=\frac{10,000}{1,000,000}[/tex]

[tex]A=0.01\ m^2[/tex]

The largest surface area window that can be used is [tex]\mathbf{A=0.01\ m^2}[/tex]

Answer:

false

Explanation:

there is difference in m/s and m.s

Answer:

hola me llamo bruno y tu?

Explanation:

pero yo soy de mexico

A is a n alternator device that will allow you to reduce the voltage from 120 vac to vdc in order to charge a cell phone. the correct option is option 4.

Voltage is a measure of electric potential energy every unit charge. If an electrical charge is placed at a certain spot, the voltage demonstrates the potential energies of just that charge at that location.

To put it another way, it is a measure of the energy stored inside an electric field or perhaps an electric circuit at a certain position. It is the labor required every unit charge against it electric field to transport the charge from one location to another.

Therefore, a is a n alternator device that will allow you to reduce the voltage from 120 vac to vdc in order to charge a cell phone. the correct option is option 4.

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Given :

The average acceleration of a tennis ball that has an initial velocity of 6.0 m/s.

and a final velocity of 7.3 m/s.

It is in contact with a tennis racket for 0.094 s

To Find :

The average acceleration of the tennis ball.

Solution :

We know, average acceleration is given by :

[tex]a_{avg}=\dfrac{Final \ velocity-Initial\ velocity}{Time\ Taken}\\\\a_{avg}=\dfrac{7.3-6.0}{0.094}\ m/s^2\\\\a_{avg}=13.83\ m/s^2[/tex]

Therefore, average velocity is given by 13.83 m/s².

Hence, this is the required solution.

Answer:

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

N is normal reaction = mg

Fs = μmg

μ = Fs / mg

μ = (27.4) / (73.2 x 9.8)

μ = 0.038

Therefore, the coefficient of friction in the hall is 0.038

Answer:

a. TRUE

Explanation:

When a satellite is launched to orbit around earth, it has to produce its own artificial gravity by performing rotations. The frequency of this rotation is given by the following formula:

f = √[ac/4πR²]

where,

f = frequency

ac = centripetal acceleration

R = Radius of the satellite

Therefore, it is clear from this formula that the frequency of rotation of the satellite is independent of its height above the surface of earth. So, the correct option is:

a. TRUE

Answer:

There are seven base SI units corresponding to different parameters and are considered independent of each other. Derived units are derived from these 7 base units. Derived units are dependent on the base units and are not independent of each other. For example, let us derive the SI units for force. By Newtons Second Law of Motion, the force is same as the product of mass and acceleration of a body. that is, force = mass x acceleration. Also, acceleration is defined as the rate of change of velocity. And velocity is defined as the rate of change of displacement. Thus, force = mass x acceleration = mass x dv/dt = mass x dt2 Mass has SI units of kg, distance is measured in m and t has the SI unit of second. Thus, Sl unit of force is kg.m/s^2 (also known as Newton).

a) the answer is delta waves i think

Answer:

A

Explanation:

Answer:

Usually when climbing, it's best to be in the small front ring and the largest back ring. If your cadence is about 100 rpm, then whatever gear you're in is fine. It depends on the road, but as long as your pedaling is at a level you're comfortable with, you're fine.

Explanation:

Google answer by the way.

Answer:

they are :

Explanation:

Answer:

Buoyant force is directly proportional to the density of the fluid in which an object is immersed. Buoyancy is the tendency to rise or float in a fluid. The upward force exerted on objects submerged in fluids is called the buoyant force. where ρ is the density, V is the volume, and m is the mass of the displaced fluid.

Explanation:

Answer:

well It means the plant cells inside of a organism

Given :

A rectangular solid box of aluminum is heated.

To Find :

Which of the following is not true.

A) Its mass remains constant

C) its density increase

B) Its volume increase

D) none of the above​

Solution :

We know, when an object is heated it expands.

Objects expands means that the volume increases but mass remains same by conservation of mass.

Since mass remains constant and volume increases.

So, density will decrease.

Hence, this is the required solution.

Answer:

Explanation:

The power of the crane can be found by using the formula

[tex]p = \frac{f \times d}{t} \\ [/tex]

f is the Force

d is the distance

t is the time taken

From the question we have

[tex]p = \frac{2400 \times 10}{15} = \frac{24000}{15} \\ [/tex]

We have

power = 1600 W

But 1 W = 0.0013 hp

So 1600 = 2.08 hp

We have the final answer as

Hope this helps you

Answer:

Celsius. i think! I might be wrong

Answer:

ΔT = 4.9°C

Explanation:

The thermal energy of the bar can be given as follows:

Thermal Energy = mCΔT

where,

m = mass of bar = 1 kg

C = specific heat capacity of aluminum = 1020 J/kg.°C

ΔT = Change in Temperature = ?

Therefore,

5000 J = (1 kg)(1020 J/kg.°C)ΔT

ΔT = (5000 J)/(1020 J/°C)

ΔT = 4.9°C

The two fundamental forces that are only attractive is  A. Gravitational and strong nuclear.

The "strong" nuclear force outperforms gravity by a factor of 10 to the 38th. While gravitational and electromagnetic forces have relatively large ranges of action, the strong and weak nuclear forces operate at close ranges. All objects are affected by the strong and weak nuclear forces, whereas smaller objects are affected by gravitational and electromagnetic forces.

Nuclear forces are the most enticing of the four fundamental forces. There was no explanation provided as to how the nucleus in the atom is kept together by electromagnetism, which maintains matter together.

Therefore, option A is correct.

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Answer:

[tex]v_m \approx -4.38\; \rm m \cdot s^{-1}[/tex] (moving toward the incline.)

[tex]v_M \approx 4.02\; \rm m \cdot s^{-1}[/tex] (moving away from the incline.)

(Assumption: [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)

Explanation:

If [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex], the potential energy of the block of [tex]m = 2.20\; \rm kg[/tex] would be [tex]m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J[/tex] when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this [tex]m = 2.20\; \rm kg\![/tex] block right before the collision would also be approximately [tex]77.695\; \rm J[/tex].

Calculate the velocity of that [tex]m = 2.20\; \rm kg[/tex] based on its kinetic energy:

[tex]\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}[/tex].

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

[tex]p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}[/tex].

[tex]p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}[/tex].

Sum of the momentum of each block right before the collision: approximately [tex]18.489\; \rm kg \cdot m \cdot s^{-1}[/tex].

Sum of the momentum of each block right after the collision: [tex](m\cdot v_m + m \cdot v_M)[/tex].

For momentum to conserve in this collision, [tex]v_m[/tex] and [tex]v_M[/tex] should ensure that [tex]m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}[/tex].

Kinetic energy of the two blocks right before the collision: approximately [tex]77.695\; \rm J[/tex] and [tex]0\; \rm J[/tex]. Sum of these two values: approximately [tex]77.695\; \rm J\![/tex].

Sum of the energy of each block right after the collision:

[tex]\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)[/tex].

Similarly, for kinetic energy to conserve in this collision, [tex]v_m[/tex] and [tex]v_M[/tex] should ensure that [tex]\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J[/tex].

Combine to obtain two equations about [tex]v_m[/tex] and [tex]v_M[/tex] (given that [tex]m = 2.20\; \rm kg[/tex] whereas [tex]M = 7.00\; \rm kg[/tex].)

[tex]\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.[/tex].

Solve for [tex]v_m[/tex] and [tex]v_M[/tex] (ignore the root where [tex]v_M = 0[/tex].)

[tex]\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.[/tex].

The collision flipped the sign of the velocity of the [tex]m = 2.20\; \rm kg[/tex] block. In other words, this block is moving backwards towards the incline after the collision.