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4 years ago

A 3000 n force acts on a 200 kg object what is the acceleration of the object

Physics

2 answers:

lesantik [10]4 years ago

7 0

F=ma. a= F/m = 3000/200 = 15 m/s

Virty [35]4 years ago

3 0

Explanation:

As we know that relation between force, mass and acceleration is as follows.

F = ma

where, F = force

m = mass

a = acceleration

So, it is given that force is 3000 N and mass is 200 kg. Hence, calculate the acceleration as follows.

F = ma

3000 N = $200 \times a$

a = $\frac{3000}{200}$

= $15 m/s^{2}$

Thus, we can conclude that acceleration of the given object is $15 m/s^{2}$ .

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An aluminum wire with a diameter of 0.100mm has a uniform electric field of 0.200V/m imposed along its entire length. The temper

Aleonysh [2.5K]

The linear resistivity of this wire is equal to 3.15 × 10⁻⁸ Ωm.
The current density in this wire is equal to 6.35 × 10⁶ A/m².
The total current in a wire is equal to 0.0499 Amp.
The drift speed of the conduction electrons is equal to 6.59 × 10⁻⁴ m/s.
The potential difference between the ends of this wire is equal to 0.4 Volt.

<u>Given the following data:</u>

Diameter of aluminum wire = 0.100 mm.

Uniform electric field of aluminum wire = 0.200 V/m.

Temperature of aluminum wire = 50.0°C.

<u>Scientific data:</u>

Resistivity of aluminum, ρ = 2.82 × 10⁻⁸ Ωm

Temperature coefficient for aluminum, α = 3.9 × 10⁻³ °C⁻¹.

<h3>How to determine the resistivity?</h3>

Mathematically, the linear resistivity of a material can be calculated by using this formula:

ρ = ρ₀(1 + αΔT)

ρ = ρ₀(1 + α(T₂ - T₁)

ρ = 2.82 × 10⁻⁸ × [1 + 3.9 × 10⁻³(50 - 20)

Resistivity, ρ = 3.15 × 10⁻⁸ Ωm.

<h3>What is the current density in this wire?</h3>

Mathematically, the current density in a wire can be calculated by using this formula:

J = σE = E/ρ

J = 0.2/3.15 × 10⁻⁸

Current density, J = 6.35 × 10⁶ A/m².

<h3>What is the total current in this wire?</h3>

Mathematically, the total current in a wire can be calculated by using this formula:

I = JA = J(πr²)

I = 6.35 × 10⁶ × (3.142 × 0.00005²)

Total current, I = 0.0499 Amp.

<h3>What is the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = I/nqA

V = (0.0499 × 0.027)/(6.023 × 10²³ × 27000 × 1.602 × 10⁻¹⁹ × (3.142 × 0.00005²)

Drift speed, V = 6.59 × 10⁻⁴ m/s.

For the the potential difference, we have:

Mathematically, the potential difference between the ends of a wire can be calculated by using this formula:

ΔV = El

ΔV = 0.2 × 2

ΔV = 0.4 Volt.

Read more on drift speed here: brainly.com/question/15219891

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Complete Question:

An aluminum wire with a diameter of 0.100 mm has a uniform electric field of 0.200 V/m imposed along its entire length. The temperature of the wire is 50.0°C. Assume one free electron per atom.

(a) Determine the resistivity.

(b) What is the current density in the wire?

(d) What is the drift speed of the conduction electrons?

(e) What potential difference must exist between the ends of a 2.00-m length of the wire to produce the stated electric field?

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2 years ago

A 1.0 104 kg spacecraft is traveling through space with a speed of 1200 m/s relative to Earth. A thruster fires for 2.0 min, exe

aniked [119]

We are given information:
$m=1.0* 10^{4} kg \\ v=1200m/s \\ t=2min=120s \\ F = 25kN = 25000N$

If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2

Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s

Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s

Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s

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3 years ago

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

Svetllana [295]

Answer:

$P_G = 14.03 \ psig$

$h_m = 0.148 \ m$

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is $P_{m} = 15.5 \ psig = 15.5 * 6894.76 = 106868.78 \ N/m^2$

The level of mercury is $h = 950 \ mm = 0.950 \ m$

The drop in the mercury level at the visible arm is $d = 39.0 = 0.039 \ m$

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

$P_g = P_m = g * \delta h * \rho$

Here $\rho$ is the density of mercury with value $\rho = 13.6 *10^{3} kg/m^3$

and $\delta h$ is the difference in the level of gas in arm one and two

$\delta h = \frac{106868.78}{ 13.6 *10^{3} * 9.8 }$

$\delta h = 0.802 \ m$

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

$h_m = 0.950 - 0.802$

=> $h_m = 0.148 \ m$

Generally from manometry principle we have that

$P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0$

Here $P_G$ is the pressure of the gas

$P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0$

$P_G = 9.6724 04 *10^{4} \ N/m^2$

converting to psig

$P_G = \frac{ 9.6724 04 *10^{4} }{6894.76}$

$P_G = 14.03 \ psig$

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3 years ago

What is the length of the blues progression in this excerpt?

Gre4nikov [31]

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3 years ago

Q.4. What is the kinetic energy of a 10 kg car that is moving 4 m/s?

kodGreya [7K]

An object of mass $10 \mathrm{~kg}$ is moving with a uniform velocity of $4 \mathrm{~ms}^{-1}$ . The kinetic energy possessed by the object is $80 \mathrm{~J}$ .

Given:

Mass of an object $=10 \mathrm{~kg}$

Velocity $=4 \mathrm{~ms}^{-1}$

Kinetic Energy $=1 / 2 \times$ Mass of Object $\times(\text { Velocity })^{2}$

$\Rightarrow$ Kinetic Energy $=1 / 2 \times 10 \times 4 \times 4$

$\Rightarrow$ Kinetic Energy $=\underline{80 \mathbf{J}}$

What is Kinetic Energy?

In physics, an object's kinetic energy is the energy it has as a result of its motion.
It is defined as the amount of work required to accelerate a body of a given mass from rest to a certain velocity.
The body retains its kinetic energy after gaining it during acceleration until its speed changes.
Kinetic energy is present in a speeding bullet, a walking human, and electromagnetic radiation such as light. The energy associated with the continual, random bouncing of atoms or molecules is another type of kinetic energy.

Learn more about kinetic energy brainly.com/question/12669551

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