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3 years ago

What is the resistance of a 1.3-m-long copper wire that is 0.30 mm in diameter?

1 answer:

4 0

The resistance R of a wire is given by:
$R= \frac{\rho L}{A}$
where
$\rho$ is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire.

For the wire in the problem, the resistivity is (copper resistivity) $\rho=1.68 \cdot 10^{-8} \Omega m$ . The length of the wire is L=1.3 m, while the cross-sectional area is
$A=\pi r^2 = \pi ( \frac{d}{2})^2 = \pi ( \frac{0.30 \cdot 10^{-3} m}{2} )^2 =7.07 \cdot 10^{-8} m^2$

so the resistance of the wire is:
$R= \frac{(1.68 \cdot 10^{-8} \Omega m)(1.3 m)}{7.07 \cdot 10^{-8} m^2}=0.31 \Omega$

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What law states force is dependent on the mass and acceleration of an object

UNO [17]

Answer:

Newton's second law of motion

Explanation:

Newton's second law of motion can be stated

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

in another form,

Force = mass * acceleration

5 0

3 years ago

A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits

Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

5 0

3 years ago

QUICK ITS DUE IN 6 MINUTES. create a scenario that applies the 3 laws of motion. Explain in complete sentences how each law if d

yKpoI14uk [10]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

5 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago

6. A 4 kg object hangs below a 6 kg object by a string of negligible mass. If the 6 kg object is pulled upward by a force of 440

MrRissso [65]

Answer:

T =176 N

Explanation:

from diagram

F -(m_1+m_2_g) = (m_1+m_2_g)a

440 - (6+4)g = (6+4)a

a =\frac{440-10*9.8}{10}

a =34.2 m/s^2

frrom free body diagram of mass m2 = 4kg

T -m_2g =m_2a

T = m_2(g +a)

T = 4(9.81+34.2)

T =176 N

7 0

3 years ago