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son4ous [18]
3 years ago
9

Please help me**** i need some answers now

Physics
1 answer:
sweet-ann [11.9K]3 years ago
4 0
Output can not be greater than input because the conversion of energy can not be greater than 100%.
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A plane flies 1800 miles in 9 ​hours, with a tailwind all the way. the return trip on the same​ route, now with a​ headwind, tak
Fittoniya [83]

Initially its moving with tail wind so here the speed of wind will support the motion of the plane

so we can say

V_{plane} + v_{wind} = \frac{distance}{time}

V_{plane} + v_{wind} = \frac{1800}{9}

V_{plane} + v_{wind} = 200 mph

now when its moving with head wind we can say that wind is opposite to the motion of the plane

V_{plane} - v_{wind} = \frac{distance}{time}

V_{plane} - v_{wind} = \frac{1800}{12}

V_{plane} - v_{wind} = 150mph

now by using above two equations we can find speed of palne as well as speed of wind

V_{plane} = 175 mph

v_{wind} = 25 mph

5 0
4 years ago
check the correctness of physical equation f is equals to MV square upon R where f is the centripetal force acting on a body of
Bas_tet [7]

Answer:

We know ForceF=mass×acceleration----------(1)

And given ForceF=

r

mv

2

----------(2)

Equating 1 and 2 dimensionaly we get ,

ma=

r

mv

2

MLT

−2

=ML

2

T

−2

L

−1

MLT

−2

=MLT

−2

8 0
3 years ago
If your car tachometer says your engine is moving at 1200 RPM, then what is it's angular velocity in Rad/sec?
Ostrovityanka [42]

Answer:

125.66 R/s

Explanation:

First    1200  r / min = 20 r/sec

20 r/s  *  2pi Radians / r = 40 pi Radians / sec = 125.66 R/s

3 0
2 years ago
If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?
Arada [10]

The force on the fry is 0.015 N

Explanation:

We can find the force acting on the fry by using Newton's second law:

F=ma

where

F is the net force on the fry

m is its mass

a is its acceleration

For the fry in this problem,

m=3 g = 0.003 kg

a=5.0 m/s^2

Therefore, the force exerted on the fry is

F=(0.003)(5)=0.015 N

Learn more about Newton's second law here:

brainly.com/question/3820012

#LearnwithBrainly

3 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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