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MrRa [10]
3 years ago
9

Autobus porusza się ze stałą prędkością 54 km/h Oblicz w jakim czasie przebędzie drogę 450 m

Physics
1 answer:
maria [59]3 years ago
3 0

Answer:

Autobus porusza się ze stałą prędkością 54 km/h Oblicz w jakim czasie przebędzie drogę 450 m

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netineya [11]

Answer:

600 J

Explanation:

it's obviously btw so yeahhhh

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3 years ago
How to light prodeced by nature
frutty [35]
Natural sources of light include our sun and other stars, where the source of energy is nuclear energy (recall that the moon does not produce light but merely reflects sunlight), lightning, where the source is electrical, and fire, where the energy source is chemical.
4 0
3 years ago
(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle. How many revolutions does the object
Travka [436]

Answer: 10.34

Explanation:

Given

\omega -t graph for a particle is given

angle turned by the particle in radians is given by the area under \omega -t graph

The area is given by

A=20\times (2-0)+10(4-2)+\dfrac{1}{2}\times (20-10)\times (3-2)\\A=40+20+5=65\ rad

Revolutions(N) made by the object is given by

N=\dfrac{65}{2\pi }=10.34

4 0
2 years ago
Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien
Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = \mu *m*g   .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5  N

F = 86.5 N

so

Acceleration in the First 15 second  will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be  as

v = u + at   ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S  in 15 s

s = u × t + 0.5 × a × t²  

here u is 0

so distance s will be

s = 0.5 × a × t²  

s = 0.5 ×  1.15 × 15²  

s = 129.37 m

and

now  acceleration acting is

F =  \mu *m*g  

m a =  \mu *m*g

a = \mu* g

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s    

here v velocity is 0  and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s    

solve it we get

s = 151.82 m

so

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

7 0
3 years ago
A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

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

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

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0
3 years ago
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