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gulaghasi [49]
3 years ago
10

Freeee points :)))))

Physics
2 answers:
zlopas [31]3 years ago
7 0

Answer:

Thank you for the free points

Explanation:

pls mark me brainly too

PLS:)

oee [108]3 years ago
6 0

Answer:

only 5 tho lol jk thx tysm

Explanation:

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A Lamborghini Aventador engine reaches its top speed from rest in 3 seconds. The engine performs 1,545,000 Joules of work in tha
nikitadnepr [17]

Answer:

W = 1.545E6 J   total work

P = W / t = 1.545E6 J / 3 sec = 5.15E5 J/sec = 515,000 J/sec  (Watts)

Using definition of power

5 0
2 years ago
An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the
Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0
3 years ago
Which of the following biomes might be found at a latitude of 33° South?
cupoosta [38]
The answer is going be desert. 
5 0
3 years ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
4 years ago
A 0.90-kg falcon is diving at 28.0 m/s at a downward angle of 35° . It catches a 0.325-kg pigeon from behind in midair. What is
pashok25 [27]

Answer:

22.11 m / s

Explanation:

The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )

initial velocity of falcon = 28 cos 35 i - 28 sin 35 j  

( falcon was flying in south east direction making 35 degree from the east )

momentum = .9 ( 28 cos 35 i - 28 sin 35 j  )

= 20.64 i - 14.45 j

initial velocity of pigeon

= 7 i

initial momentum = .325 x 7i

= 2.275 i

If final velocity of composite mass of falcon and pigeon be V

Applying law of conservation of momentum

( .9 + .325) V = 20.64 i - 14.45 j +2.275 i

V = ( 22.915 i - 14.45 j ) / 1.225

= 18.70 i - 11.8 j

magnitude of V

= √ [  (18.7 )² + ( 11.8 )²]

= 22.11 m / s

6 0
4 years ago
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