A mass of 5.00 kg in a gravitational field of 7.00 n/kg. What force acts on the mass?

La densidad del metal como se requiere en la pregunta es 7.8 * 10 ^ 3 Kg / m ^ 3.

Sabemos que ese empuje hacia arriba = peso en aire - peso en líquido

Peso en el aire = 63,5 * 10 ^ -3 Kg * 10 m / s ^ 2 = 0,635 N

Peso en líquido = 55,4 * 10 ^ -3 Kg * 10m / s ^ 2 = 0,554 N

Empuje hacia arriba = 0,635 N - 0,554 N = 0,081 N

Pero ;

Empuje hacia arriba = Volumen * densidad del fluido * aceleración debido a la gratitud

volumen = empuje hacia arriba / densidad del fluido * aceleración debido a la gratitud

volumen = 0.081 N / 1000 * 10

Volumen = 8.1 * 10 ^ -6 m ^ 3

Densidad = masa / volumen

Densidad = 63,5 * 10 ^ -3 Kg / 8,1 * 10 ^ -6 m ^ 3

= 7,8 * 10 ^ 3 kg / m ^ 3

Más información: https://brainly.com/question/12108425

Answer:

B. Radio waves

Explanation:

Radio waves are transmitted easily through air. They do not cause damage if absorbed by the human body, and they can be reflected to change their direction. These properties make them ideal for communications.

Hope this helps!!

Answer:

B

Explanation:

Radio Waves

Answer:

We can only find the change in velocity by the area under the acceleration time graph by calculating the area under the curve to the given time. Then by the formula of the difference in the final and initial velocity you can find the final velocity of the object in the given time period.

Answer:

it states that energy can either be gained or lost but it only changes its form.

Explanation:

for example:as a ball is still on the table it posses a potential energy of 100j and a k.e of 0j,as it falls it gains k.e so the midpoint the p.e is equal to the k.e (50j equally) as it approches the ground it completely gains k.e (100j) and the p.e is 0j.

total energy is 100j so it has been converted from p.e to k.e.

hope u have understood.

Answer: Increasing the amount of sugar.

Explanation: Increasing the amount of reactants increases the rate of the reaction. This is given by the collision theory, which states that a chemical reaction must involve the collisions of particles together, as well as with the walls of the container it is in. By increasing the amount of reactant in the container you are increasing the collision frequency, and the rate of the reaction is dependant on the collision frequency multiplied by the fraction of collisions that are effective. Increasing the temperature increases the collision frequency as well as the fractions of collisions that are effective, so it’s definitely not that one. Hope this helped:)

The characteristics of the speed of the waves in strings and the resonance allows to find the change in the fundamental frequency when changing the tension is:

The speed of the chord wave is given by the relationship between the tension and the density of the medium.

          [tex]v= \sqrt{\frac{T}{\mu } }[/tex]  

Where v is the velocity of the wave, T the tension of the string and μ the density

In a rope held at the ends, a process of standing waves occurs, two at the point where it is attached we have a node and a anti-node in the center.

             2L = n λ

Where L is the length of the chord and call the wavelength

Wave speeds are related to wavelength and frequency.

        v = λ f

We substitute.

            [tex]\sqrt{\frac{T}{\mu } } = \frac{2L}{n} \ \ f[/tex]  

For the fundamental frequency n = 1

            f₀ = [tex]f_o = \sqrt{\frac{T}{\mu } } \ \ \frac{1}{2L}[/tex]  

They indicate that the tension increases 1.70%

           T = T₀ + 0.17 T₀

           T = 1.17 T₀

We substitute.

         [tex]f = \sqrt{1.17 } \ \sqrt{\frac{T_o}{\mu } } \ \ \frac{1}{2L}[/tex]

         f = ra1.17 f₀

         f = 1.08 f₀

In conclusion, using the characteristics of the velocity of the waves in strings and the resonance we can find the change in the fundamental frequency when changing the tension is:

Learn more about string resonance here: brainly.com/question/16010929

Answer:

1. A and C are balanced forces as all the forces get cancelled by the opposite and equal force.

2. B and D as in B the Net force is 5N to the right and in D the Net force is 20N to the right.

3. This would be all the unbalance forces which are B and D

4. B would be moving 5N to the right and D would move 20N to the right.

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

Answer:

Approximately [tex]106\; \rm m[/tex], [tex](1/16)[/tex] of the original distance, assuming that the acceleration is constant.

Explanation:

If the acceleration on this vehicle is constant, the SUVAT equations would apply:

[tex]\displaystyle v^{2} - u^{2} = 2\, a\, x[/tex].

Rearrange this equation to find an expression for [tex]x[/tex], the displacement required for the vehicle to come to a stop:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{-u^{2}}{2\, a} && (\text{$v = 0\; \rm m \cdot s^{-1}$})\end{aligned}[/tex].

Thus, under these assumptions, [tex]x[/tex] would be proportional to [tex]u^{2}[/tex]. In other words, the distance required for this vehicle to come to a stop would be proportional to the square of the initial velocity of this vehicle.

If the initial velocity [tex]u[/tex] is reduced to [tex](1/4)[/tex] (a quarter) of the initial value, the distance required for stopping this vehicle would be [tex](1/4)^{2} = (1/16)[/tex] of the initial value:

[tex]\begin{aligned}1,\!697\; {\rm m} \times \frac{1}{16} \approx 106\; \rm m\end{aligned}[/tex].

Answer:

Proton

Explanation:

The nucleus has an overall positive charge as it contains the protons. Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons.

Answer:

Answer is below….

Explanation:

The velocity is represented by the slope of a position-time graph. The faster the motion changes, the steeper the slope. The change in position divided by the corresponding change in time may be used to compute average velocity from a position-time graph.

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