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2 years ago

The velocity of an object with mass = 2kg is given as a function of time:

Physics

1 answer:

Musya8 [376]2 years ago

8 0

Answer:

The force acting on the object at $t = 2\,s$ is $\vec F = (4, 32)\,[N]$ .

Explanation:

Given that object has a constant mass in time, the force acting on the object ( $\vec F$ ), in newtons, is defined by following expression:

$\vec F = m\cdot \vec a$ (1)

Where:

$m$ - Mass, in kilograms.

$\vec a$ - Acceleration, in meters per square second.

By definition of acceleration, we know that:

$\vec a = \frac{d}{dt} \vec v$ (2)

Let suppose that given vector velocity is expressed in meters per second. If we know that $m = 2\,kg$ , $\vec v = (2\cdot t, 4\cdot t^{2})\,\left[\frac{m}{s} \right]$ and $t = 2\,s$ , then the force acting on the object is:

$\vec a = (2, 8\cdot t)\,\left[\frac{m}{s^{2}} \right]$

$\vec F = (4, 16\cdot t)\,[N]$

$\vec F = (4, 32)\,[N]$

The force acting on the object at $t = 2\,s$ is $\vec F = (4, 32)\,[N]$ .

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What is the length of an aluminum rod at 65Â°C if its length at 15Â°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.2

Deffense [45]

Answer:

Option B is the correct answer.

Explanation:

Thermal expansion

$\Delta L=L\alpha \Delta T$

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

$\Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m$

New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m

Option B is the correct answer.

3 0

3 years ago

Read 2 more answers

Is Natural Gas nonrenewable or renewable? Why? Use in your own words.

Nutka1998 [239]

Answer:

Natural gas is non renewable energy.

Explanation:

Because they were formed from the buried reamains of plants and animals that live a million years ago. It is formed from fossil fuels.

6 0

2 years ago

A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

Tanzania [10]

Answer:

The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

Explanation:

Given that,

Radius = 5 cm

Charge density $J= 400\ nC/m^3$

We need to calculate the total charge Q of the sphere

Using formula of charge

$q=\rho V$

Where, $\rho$ = charge density

V = volume

Put the value into the formula

$q=\rho\times(\dfrac{4}{3}\pi r^3)$

Put the value into the formula

$q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3$

$q=2.094\times10^{-10}\ C$

Hence, The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

6 0

2 years ago

A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

It requires 3480 J to move a 675 N object how far?

goblinko [34]

5.16 meters!!!!!!!!!!!!!!!!!!!!

4 0

2 years ago