diagram of an atom with labels

Answer:

The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.

Answer:

Explanation:

The mass is essentially "how much stuff" is in an object. ... Weight: There is a gravitational interaction between objects that have mass. If you consider an object interacting with the Earth, this force is called the weight. The unit for weight is the Newton (same as for any other force).

Answer:

If it accelerates at 20 m/s2 for a period of 22 seconds, what is its final velocity? ... How fast is the ball falling after 5 seconds? v = v0 + gt v = 0 + 10(5) v = 50 m/s. 4. ... + ½ 2.5(15)2 x = 281 m. 5. What is the total displacement of the car in question 2? ... 8. A base jumper falls until he reaches a speed of 200 m/s

Explanation:

Answer:

No, it is very unlikely to perform a controlled experiment, because you need to observe the amount or anything from something. Consider someone on the busy street of a New York neighborhood asking random people that pass by how many pets they have, then taking this data and using it to decide if there should be more pet food stores in that area.

Answer:−4.05

Explanation:

Answer:

The net acceleration of the SUV is 0.429 meters per square second due west.

Explanation:

Statement is incomplete. Description is presented below:

A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?

According to Newton's Laws of Motion, the SUV will accelerate if and only if net acceleration is different of zero. Let suppose as positive the direction of driving force ([tex]F[/tex]), measured in newtons:

[tex]\Sigma F = F - R -f = F_{net}[/tex] (1)

Where:

[tex]R[/tex] - Resistance force, measured in newtons.

[tex]f[/tex] - Wind force, measured in newtons.

[tex]F_{net}[/tex] - SUV net force, measured in newtons.

If we know that [tex]F = 2500\,N[/tex], [tex]R = 500\,N[/tex] and [tex]f = 500\,N[/tex], then net force experimented by the SUV is:

[tex]F_{net} = 2500\,N-500\,N-500\,N[/tex]

[tex]F_{net} = 1500\,N[/tex]

The car has acceleration.

By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:

[tex]a_{net} = \frac{F_{net}}{m}[/tex] (2)

Where [tex]m[/tex] is the mass of the SUV, measured in kilograms.

If we know that [tex]F_{net} = 1500\,N[/tex] and [tex]m = 3500\,kg[/tex], then the net acceleration of the car is:

[tex]a_{net} = \frac{1500\,N}{3500\,kg}[/tex]

[tex]a_{net} = 0.429\,\frac{m}{s^{2}}[/tex]

The net acceleration of the SUV is 0.429 meters per square second due west.

Answer:

el conductor

Explanation:

gracias por los puntitoss

Answer:

conductor

Explanation:

Answer:

63°

Explanation:

90-27 =63

I am not completely sure

Answer:

a.432

B. 562

C.402

D. 316

A

Explanation:

Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied. Work was done by the workman will be 432 Nm.

What is work?

Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative.it depend on the diection of body displaced . if the body is displaced in thw same direction of force it will be positive .

while if the displacement is in the opposite direction of force applied the work will be negative work . if their is no displacement of the body the work done will b e zero.

As given in question wall is stationary diplacement will be zero so in that work will also z

The given data in the problem is;

x is the displacement in the horizontal direection=22m

y is the displacement in the verical direection=9.6m

W is the weight of lumber = 45 N

[tex]\rm W=F\times y[/tex]

[tex]\rm W=F\times y \\\\\rm W=45\times 9.6\\\\ \rm W=432\;Nm[/tex]

Hence Work was done by the workman will be 432 Nm.

To learn more about the work refer to the link ;

https://brainly.com/question/3902440

Answer:

She should explain that the Sun is made up of gaseous layers that surround an iron core.

Answer:

7.22 × 10²⁹ kg

Explanation:

For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.

So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

M = 7.21766 × 10²⁹ kg

M ≅ 7.22 × 10²⁹ kg

Answer:

Um its the vbuck card on the 3 thrid level

Explanation:

Bc its a vbuck card you know sihdg;aig

Answer:

Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.

Explanation:

Answer:

a= 0

Explanation:

       [tex]F_{net} = m*a = 0 (1)[/tex]

      ⇒ a = 0

Answer: Terraces on moderate to steep irregular slopes pro- ... sure of infertile or toxic soils. ... Following are terms used to define distances mea- ... the soil in the entire field will be disturbed to con-.

Explanation:

Answer:

A: 1,212 kg

Explanation:

Answer:

3.6 arcsec

Explanation:

 angular diameter = diameter / distance

diameter is constant

so angular diameter ∝ 1 / distance

angular diameter = k / distance

For first case ,

18 = k / .5

for second case let angular diameter be D .

D = k / 2.5

dividing ,

D / 18 = .5 / 2.5 = 1 / 5

D = 18 / 5 = 3.6 arcsec

3.6 arcsec is the answer .

Answer:

d=20m/sx60s=1200m=1200/1000Km=1.2km

Explanation:

Answer:

A. [tex]2\cdot t[/tex]

Explanation:

The SUV accelerates uniformly. In this case, speed ([tex]v[/tex]) is directly proportional to time ([tex]t[/tex]). That is:

[tex]v \propto t[/tex] (1)

[tex]v = k\cdot t[/tex]

Where [tex]k[/tex] is the proportionality constant.

Then, we eliminate this constant by creating the following relationship:

[tex]\frac{v_{1}}{t_{1}} = \frac{v_{2}}{t_{2}}[/tex] (2)

If we know that [tex]v_{1} = v[/tex], [tex]t_{1} = t[/tex] and [tex]v_{2} = 2\cdot v[/tex], then the time interval for the SUV to accelerate from rest to a speed [tex]2\cdot v[/tex] is:

[tex]\frac{v}{t} = \frac{2\cdot v}{t_{2} }[/tex]

[tex]t_{2} = 2\cdot t[/tex]

Hence, correct answer is A.

Answer:

The answer is below

Explanation:

The resistance of a wire is directly proportional to the length of the wire and inversely proportional to its area. The resistance (R) is given by:

[tex]R=\frac{\rho L}{A}\\\\where\ L=length \ of\ wire,A=cross\ sectional\ area, \rho=resistivity\ of\ wire.[/tex]

Let us assume that all the wires have the same resistivity.

a) Wire of Length L and area A

[tex]R_1=\frac{\rho L}{A}[/tex]

b) Wire of Length 2L and area A

[tex]R_2=\frac{\rho *2L}{A}=2R_1[/tex]

C) Wire of Length L and area 2A

[tex]R_3=\frac{\rho L}{2A}=\frac{1}{2}R_1[/tex]

Therefore the wire of least resistance is R3 and R2 has the highest resistivity.

R₃ < R₁ < R₂

Therefore, the ranking of the wires from most current (least resistance) to least current (most resistance) is:

R₃ < R₁ < R₂

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Answer:

At zero kelvin (minus 273 degrees Celsius) the particles stop moving and all disorder disappears. Thus, nothing can be colder than absolute zero on the Kelvin scale. Physicists have now created an atomic gas in the laboratory that nonetheless has negative Kelvin values.

Explanation:

are you a dmbss? the bulb and wire must be connected to both end

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          [tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_i m_i[/tex]

in this case we have two bodies

          x_{cm} = [tex]\frac{1}{M}[/tex] (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = [tex]\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )[/tex]

          x_{cm} = [tex]\frac{1}{M} ( m_1 v_1 + m_2 v_2 )[/tex]

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = [tex]\frac{m_1}{m_1+m_2}[/tex]  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         [tex]v_{cm o}[/tex] = [tex]\frac{1}{m_1 +m_2}[/tex] (m₁ vo)

final

         [tex]v_{cm f}[/tex] = [tex]\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )[/tex] ( v) = v

         v_{cm f} =  [tex]\frac{m_1}{M^2} v_o[/tex]

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = [tex]\frac{1}{M} = \frac{1}{m_1 +m_2}[/tex]

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = [tex]\frac{m_1 -m_2}{m1 +m2 } v_o[/tex]

             v_{2f} = [tex]\frac{2 m_1}{m-1+m_2} vo[/tex]

now let's find the velocity of the center of mass

initial

          [tex]v_{cm o}[/tex] = [tex]\frac{1}{M}[/tex] m₁ v₀

final

          [tex]v_{cm f}[/tex] = [tex]\frac{1}{M}[/tex]  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = [tex]\frac{1}{M}[/tex] [  [tex]m_1 \frac{m_2}{M}[/tex] + [tex]m_2 \frac{2 m_1}{M}[/tex] ] v₀

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = [tex]\frac{1}{M}[/tex] M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

Answer:

a) the rate at which 235U is fissioned is 2895.9 grams per day

b) the rate at which 235U is consumed is 3385.3071 gram/day

Explanation:

Given the data in the question;

a)

designed operation thermal power = 2758 MW.

we know that, the burn up rate fission rate of 235U is 1.05 grams per MW-day.

so, the rate at which 235U is fissioned will be;

⇒ 2758 × 1.05 = 2895.9 grams per day

Therefore, the rate at which 235U is fissioned is 2895.9 grams per day

b)

Consumption rate is given as;

Cr = 1.05 × ( 1 + ∝ )P gram/day

where ∝ is 0.169 for U-235

so,

Cr = 1.05 × ( 1 + 0.169 )2758 gram/day

Cr = 1.05 × 1.169 × 2758 gram/day

Cr = 3385.3071 gram/day

Therefore, the rate at which 235U is consumed is 3385.3071 gram/day

Answer:

A

Explanation:

In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.

Hope this helps!

Answer:

[tex]c=0.45\ J/g^{\circ} C[/tex]

Explanation:

Given that,

A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at  30.00°C). The final temperature of the system is 40.22°C.

We need to find the specific heat of iron.

It can be calculated as:

Cooler water gains = hot metal loses

mc∆T = - mc∆T

Put all the values,

[tex]200g(4.184\ J/g^{\circ} C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C[/tex]

So, the specific heat of iron is [tex]0.45\ J/g^{\circ} C[/tex]

Answer:

she should write about how big is it and what the sun looks and how far away is it from earth.