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MAVERICK [17]
3 years ago
15

A helicopter is dropping a package of rescue supplies to a group of stranded hikers. If it drops the supplies while it is hoveri

ng 172m above the hikers, how long does it take to hit the ground?
Physics
1 answer:
Ann [662]3 years ago
5 0

Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of the average velocity at which the supplies are travelling in air. The formula to be used here if the completed question is obtained is

Average velocity (in m/s) =  distance (m) ÷ time (s)

Since we are looking for the time, then time should be made the subject of the formula

time (in secs) = distance ÷ average velocity

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5. Alex goes cruising on his dirt bike. He rides 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south 300 m east and f
Gala2k [10]

Answer:

Explanation:

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5 0
3 years ago
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
ivann1987 [24]

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

\tau=F\times r

\tau=82.0\times0.150

\tau=12.3\ N-m

Now, The angular acceleration

\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}

\dfrac{d\omega}{dt}=20.73\ rad/s^2

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

\tau=I\times\dfrac{d\omega}{dt}

I=\dfrac{I}{\dfrac{d\omega}{dt}}

I=\dfrac{12.3}{20.73}

I= 0.593\ kg-m^2

Hence, The moment of inertia of the wheel is 0.593 kg-m².

5 0
3 years ago
Which statement is not true about mass movements?
ololo11 [35]

B i think that the answer so

8 0
3 years ago
Read 2 more answers
Citez un exemple de réaction endothermiques
Fiesta28 [93]

Réponse/explication:

La réaction qui produit de l'hydrogène et de l'oxygène à partir d'eau est endothermique. Certains changements de phase comme la vaporisation, la fusion ou la sublimation et la plupart des réactions de décomposition chimique sont endothermiques.

4 0
3 years ago
A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o
IrinaK [193]

Answer:

a. 6.032\times10^{-6}C/m^2

b.6.816\times10^5N/C

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as \sigma

\sigma=\frac{Q}{4\pi r_{out}^2}, and the strength of the electric field outside the sphere E=\frac{\sigma _{new}}{\epsilon _o}

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2  #surface charge outside sphere.

\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2

Hence, the new charge density on the outside of the sphere is 6.032\times10^{-6}C/m^2

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be 6.032\times10^{-6}C/m^2,

E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C

Hence, the strength of the electric field just outside the sphere is 6.816\times10^5N/C

5 0
3 years ago
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