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katrin2010 [14]
3 years ago
12

suppose the pilot starting again from rest opens the throttle part.way at constant acceleration the airboat then covers a distan

ce of 60.0m in 10.0s find the net force action on the boat​
Physics
1 answer:
mr_godi [17]3 years ago
3 0

Answer:

Acceleration is 1.2 m/s^2.

Explanation:

initial velocity, u = 0

distance, d = 60 m

time, t = 10 s  

Let the acceleration is a.

use second equation of motion

s= u t +0.5 at^2\\\\60 =  0 + 0.5 \times a \times 10\times 10\\\\a = 1.2 m/s^2

Now according to the Newton's second law

Force = mass x acceleration

Let the mass is m.

F = m x 1.2 = 1.2 m Newton  

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Negative acceleration is also known as
STALIN [3.7K]

Answer:

Deceleration

Explanation:

The amount by which a speed or velocity decreases

8 0
3 years ago
Read 2 more answers
Two convex lenses are placed 15 cm apart. The left lens has a focal length of 10 cm, and the right lens a focal length of 5 cm.
Minchanka [31]

Answer:

The image distance from right lens is 2.86 cm and image is real.

Explanation:

Given that,

Focal length of left lens = 10 cm

Focal length of right lens = 5 cm

Distance between the lenses d= 15 cm

Object distance = 50 cm

We need to calculate the image distance from left lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-50}

\dfrac{1}{v}=\dfrac{3}{25}

v=8.33\ cm

We need to calculate the image distance from right lens

The object distance will be

u = 15-8.33 = 6.67\ cm

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{-6.67}

\dfrac{1}{v}=\dfrac{1167}{3335}

v=2.86\ cm

The image is real.

Hence, The image distance from right lens is 2.86 cm and image is real.

4 0
3 years ago
What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of
Dvinal [7]

Answer:1384 Hz

Explanation:

Given

wavelength(\lambda )=0.25 m

Temperature T=25^{\circ}

at T=25^{\circ} velocity of sound is 346 m/s

and we know

velocity=frequency\times \lambda

v=f\times \lambda

346=f\times 0.25

f=1384 Hz

5 0
3 years ago
Read 2 more answers
A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit
elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

                                            = 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

                             T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }  second

Squaring the terms and solving it for 'g'

                             g = 4 π² \frac{(R+h)^{3} }{R^{2}T^{2}  }   m/s²

Substituting the values in the above equation

                   g = 4 π² \frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}  

                                    g = 4.84 m/s²      

Therefore, the weight

                                     w = m x g   newton

                                         = 5832 Kg x 4.84 m/s²

                                         = 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N                

8 0
4 years ago
A sample container of carbon monoxide occupies a volume of 435 ml at a pressure of 785 torr and a temperature of 298 k. What wou
strojnjashka [21]

Answer:

181.54 K

Explanation:

From gas laws, we know that v1/t1= v2/t2 where v and t represent volume and temperatures, 1 and 2 for the first and second container. Making t2 the subject of the formula then

T2=v2t1/ v1

Given information

V1 435 ml

V2 265 ml

T1 298K

Substituting the given values then

T2=265*298/435=181.54 K

3 0
3 years ago
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