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4 years ago

8

The height of a helicopter above the ground is given by h = 2.80t3, where h is in meters and t is in seconds. At t = 2.45 s, the

helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

1 answer:

GalinKa [24]4 years ago

4 0

Answer:

1.45

Explanation:

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What net force is necessary to give a 2 kg mass that is initially at rest an acceleration of 5 m/s2?

QveST [7]

Answer:

<h2>10 N</h2>

Explanation:

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

force = mass × acceleration

From the question we have

force = 2 × 5

We have the final answer as

<h3>10 N</h3>

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Which parts of the roller coaster determine the amount of potential energy in the system? Explain your answer.

Paha777 [63]

Answer:

the lowest point, where the amount of energy is the greatest

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3 years ago

An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temp

tekilochka [14]

Answer with Explanation:

We are given that

Diameter of coil=d=0.115mm

Radius, r= $\frac{d}{2}=\frac{0.115}{2}=0.0575mm=0.0575\times 10^{-3} m$

Using $1mm=10^{-3} m$

Electric field=E=0.235V/m

T=55 degree C

$T_0=20^{\circ} C$

$\rho_0=2.82\times 10^{-8}\Omega m$

$\alpha=3.9\times 10^{-3}/C$

(a).We know that

$\rho=\rho_0(1+\alpha(T-T_0))$

Substitute the values

$\rho=2.82\times 10^{-8}(1+3.9\times 10^{-3}(55-20))$

$\rho=3.2\times 10^{-8}\Omega m$

(b).Current density, $J=\frac{E}{\rho}$

Using the formula

$J=\frac{0.235}{3.2\times 10^{-8}}=7.3\times 10^6A/m^2$

c.Total current,I=JA

Where $A=\pi r^2$

$\pi=3.14$

Using the formula

$I=7.3\times 10^6\times 3.14\times (0.0575\times 10^{-3})^2$

I=0.076A

d.Length of wire=l=2m

$V=El$

Substitute the values

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harkovskaia [24]

An outward force is exerted

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Answer:

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Explanation:

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3 years ago