first correct answer gets brainliest

Answer:

Its A

Explanation:

Just did the quiz

The pair of objects that should be strongly attracted is option A. A positively charged particle and a negatively charged particle.

When there is a positively charged particle  & the negatively charged particle so due to this it should be strongly attracted.  Also, when there are two positively charged particles, or a negative one or a include negative one or neutral one so it should not be strongly attracted. Therefore, the option A is correct.

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Answer:

ggggggggggggggggggggggggggggg

Explanation:

Answer:

The volume of the sample of gold is

16.51 [tex]cm^{3}[/tex]

Explanation:

The formula for density is:

D= [tex]\frac{M}{V}[/tex].

where:

D is density,

M is mass, and

V is volume.

Rearrange the density formula to isolate volume.

V= [tex]\frac{M}{D}[/tex]

V= [tex]\frac{318.97g Au}{19.32g cm^{3}}[/tex]

V= 318.97∅ ×  [tex]\frac{1 cm^{3} Au}{19.32g cm^{3} }[/tex]← Multiply by the multiplicative inverse of the density.

V= 16.51 cm³ Au.

Answer:

6.7 m/s

Explanation:

Given:

Δx = 5 m

v₀ = 5 m/s

a = 2 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (5 m/s)² + 2 (2 m/s²) (5 m)

v = 6.7 m/s

The particles of a more dense substance are closer together than the particles of a less dense substance. Thus, the given statement is true.

Density of the particles is the substance's mass per unit of volume. The symbol which is most often used for the density is ρ (rho), although the Latin letter D can also be used to denote density.

Density is the mass of a unit volume of a material substance or particle. The formula for density is d = M/V (mass per unit volume), where d is density, M is the mass of particle, and V is the volume. Density is commonly expressed in the units of grams per cubic centimeters.

The S.I. unit of density is made up of the mass of the particle which is kg and that of volume is meter cube. Hence, the S.I. unit of density is kg/m³.

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Answer:

Explanation:

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below  

Answer:

40N

Explanation:

trust

Answer:

5.0 m/s south

Hope this Helps!

Answer:

5.0 m/s south

Explanation:

Answer:

1>500gf

1>300gf

its answer

Answer:

Interviewing and observation are two methods of collecting qualitative data as part of research. ... Interviews vary from structured, in which a set list of questions is asked of every interviewee, to unstructured, which is open-ended.

Answer:

(a) 102 cm/s

(b) 0.490 cm²

Explanation:

(a) Use Bernoulli equation.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0

½ ρ v₁² + ρgh₁ = ½ ρ v₂²

½ v₁² + gh₁ = ½ v₂²

½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²

v = 102 cm/s

(b) The flow rate is constant.

v₁ A₁ = v₂ A₂

(25.0 cm/s) (2.00 cm²) = (102 cm/s) A

A = 0.490 cm²

Answer:

The value  is  [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

Explanation:

From the question we are told that  

   The speed of the airplane from the north is [tex]\frac{dN}{dt}  =  -181 \  km /hr[/tex]

The negative sign is because the direction is towards the south

  The speed of the airplane from the east is  [tex]\frac{dE}{dt}  =  -278 \  km/hr[/tex]

The negative sign is because the direction is towards the west

   The distance of the southbound plane from the airport is  [tex]N  =  30 \  km[/tex]

   The distance of the westbound plane is  [tex]E =  15 \  km[/tex]

Generally the distance between the plane is mathematically represented using Pythagoras theorem  as

    [tex]R^2  = N^2 + E^2[/tex]

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

     [tex]2R\frac{dR}{dt} =  2N \frac{dN}{dt} +   2E\frac{dE}{dt}[/tex]

Here

     [tex]R = \sqrt{N^2 + E^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R =33.54 \ m [/tex]

    [tex]2(33.54) * \frac{dR}{dt} =  2( 30)*(-181)  +   2*15*(-278)[/tex]

=>   [tex] 67.08 * \frac{dR}{dt} =  -19200[/tex]

=>   [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

The rate of change of the distance between the planes is 286.23 km/hr.

The given parameters;

The distance between the two planes is calculated by applying Pythagoras theorem as shown below;

[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]

The rate of change of the distance between the planes is calculated as follows;

[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]

Thus, the rate of change of the distance between the planes is 286.23 km/hr.

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Answer:

The moment of inertia of the object is 17.276 kilogram-square meters.

Explanation:

According to the statement, we find that object has rotation and no translation. From Rotation Physics we get that rotational kinetic energy ([tex]K_{R}[/tex]), measured in joules, is represented by the following formula:

[tex]K_{R} = \frac{1}{2}\cdot I_{G}\cdot \omega^{2}[/tex] (Eq. 1)

Where:

[tex]I_{G}[/tex] - Moment of inertia with respect to center of mass, measured in kilogram-square meters.

[tex]\omega[/tex] - Angular speed, measured in radians per second.

Now we clear the moment of inertia:

[tex]I_{G} = \frac{2\cdot K_{R}}{\omega^{2}}[/tex]

If we know that [tex]K_{R} = 16\,J[/tex] and [tex]\omega \approx 1.361\,\frac{rad}{s}[/tex], then the moment of inertia of the object is:

[tex]I_{G} = \frac{2\cdot (16\,J)}{\left(1.361\,\frac{rad}{s} \right)^{2}}[/tex]

[tex]I_{G} =17.276\,kg\cdot m^{2}[/tex]

The moment of inertia of the object is 17.276 kilogram-square meters.

The moment of inertia of the object will be "17.276 kg/m²".

Rotational Kinetic energy, [tex]K_R[/tex] = 16 J

Angular speed, ω = 1.361 rad/s

By using the Rotation Physics, the relation will be:

→ [tex]K_R[/tex] = [tex]\frac{1}{2}[/tex] × [tex]I_G[/tex] × ω²

the,

The moment of inertia be:

→ [tex]I_G[/tex] = [tex]\frac{2\times K_R}{\omega^2}[/tex]

By substituting the values, we get

       = [tex]\frac{2\times 16}{(1.361)^2}[/tex]

       = [tex]\frac{32}{(1.361)^2}[/tex]

       = 17.276 kg.m²

Thus the above answer is correct.  

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Answer:

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

Explanation:

Given that,

Constant speed = 6.38 m/s

Force [tex]F=7.50\times10^{3}\ N[/tex]

Kinetic friction = 0.26

(a). We need to calculate the friction force

Using formula of friction force

[tex]f_{k}=\mu F_{N}[/tex]

Put the value into the formula

[tex]f_{k}=0.26\times7.50\times10^{3}[/tex]

[tex]f_{k}=1950\ N[/tex]

(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,

We need to calculate the magnitude of this force

According to given data,

The same force will be applied to keep constant velocity.

Hence, (a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force.

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

a. The magnitude of the kinetic friction force experienced by the sleigh is

[tex]= 0.76 \times 7.50 \times 10^3[/tex]

= 1950 N

b. It should be equivalent to the friction force.

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Ohm's law shows the relationship between voltage, current, and the resistance of a energy bond. The formula for Ohm's law is:

voltage = current x resistance

This formula tells you that current and resistance is the voltage of an energy bond.

Hope this helps you!

Answer: What is Ohm’s Law?

Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.

Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists.

E = I × R

It means voltage = current × resistance, or volts = amps × ohms, or V = A × Ω.

Resistance cannot be measured in an operating circuit, so Ohm's Law is especially useful when it needs to be calculated. Rather than shutting off the circuit to measure resistance, a technician can determine R using the above variation of Ohm's Law.

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Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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

where

f is the force

m is the mass

From the question

f = 20 N

m = 10 kg

We have

[tex]a = \frac{20}{10} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:.

Explanation:.

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

Answer:the answer is 3

Explanation:

Answer:

there is only a small amount of gravity present.

Explanation:

this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.

Answer:

2.096m/s

Explanation:

The speed of this soccer ball can be calculated using the formula;

Speed = distance/time

According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;

Speed = 6.5/3.1

Speed of the ball = 2.096m/s

Therefore, the speed of the ball before hitting the tree is 2.096m/s

Answer:

[tex](1630.13\pm 300.10)\ kg/m^3[/tex]

Explanation:

Given that,

The radius of a sphere is (6.45 ± 0.30) cm

Mass of the sphere is (1.79 ± 0.08) kg

Density = mass/volume

For sphere,

[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]

We can find the uncertainty in volume as follows :

[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]

Uncertainty in mass,

[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]

Now, the uncertainty in density of sphere is given by :

[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]

Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]

v = u + a t

where u = initial velocity (25 m/s), v = final velocity (0), a = acceleration, and t = time (45 s). So

0 = 25 m/s + a (45 s)

a = (-25 m/s) / (45 s)

a ≈ -0.56 m/s²

This question is incomplete, the complete question is;

A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.

If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation

Answer: the time constant of the damped oscillation is 47.44s

Explanation:

Given that;

t = 5.0s

Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao

EXPRESSION for amplitude is  A(t) = Ao e^-t / T

t is time while T is time constant

so

0.9Ao = Ao e^-t / T  

0.9 = e^ -t/T

So we take the natural log of both the sides

ln (0.9) = -t/T

-0.1054 =  -t/T

0.1054 =  t/T

WE now substitute our value of t

0.1054 =  t/T

0.1054T =  5.0

T = 5 / 0.1054

T = 47.44s

therefore the time constant of the damped oscillation is 47.44s

Answer:

5400 N

Explanation:

f=ma

f= 1200*4.5

f=5400N

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

Answer:

1.3 m/s

Explanation:

It is given that,

Mass of bird A, [tex]m_A=2.2\ kg[/tex]

Mass of bird B, [tex]m_B=1.7\ kg[/tex]

Initial speed of bird A is 0 as it was at rest

Initial speed of bird B is 3 m/s

We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,

[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]

So, the center of mass for this system is 1.3 m/s.

Explanation:

because of the acceleration due to gravity is constant which is 9.8

Answer:

TRUE

Explanation:

CUZ , IT DOSEN'T AFFECT

Mass does not affect the speed of falling objects, assuming there is only gravity acting on it. Both bullets will strike the ground at the same time.