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Inessa [10]
4 years ago
14

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m

ust stop the car. If it takes 0.200 s for the driver to apply the brakes, what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign?
Physics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

The acceleration of the car will be a=9600m/sec^

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion s=ut+\frac{1}{2}at^2

200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2

a=9600m/sec^

So the acceleration of the car will be a=9600m/sec^

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Zigmanuir [339]

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

espero que te

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6 0
3 years ago
A child who is swimming toward shore at 0.78 m/s sees shark and picks up his speed
Gnoma [55]

Answer:

0.085m/s²

Explanation:

Use v²=v0²+2a(d)

solve for a

v²-v0²/2d=a

Plug in givens

1.89²-0.78²/2*17.5=a

Plug into calculator

a=0.085m/s²

3 0
3 years ago
In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o
ioda

Answer:

Explanation:

We shall apply law of conservation of  momentum in space to know the velocity of combination after the impact

m₁v₁ = m₂v₂

.1 x 4 = ( 1 + .1 ) v₂

v₂ = .3636 m /s

1  )  

Kinetic energy of the combination

= 1/2 x 1.1 x ( .3636)²

= 7.3 x 10⁻² J

2 )

Initial kinetic energy of the system

= 1/2 x 0.1 x 4²

= 0.8 J

Final  kinetic energy of the system = 7.3 x 10⁻²

Loss of energy = .8 - .073

= .727 J

This energy was converted into internal energy of the system .

3 )

increase in entropy = dQ / T

Here dQ = .727 J

T  = 300 ( Constant )

dQ / T = 2.42 X 10⁻³ J/K

6 0
4 years ago
El aspa de un ventilador de 0.30 m de diametro gira 1200 rev/min. calcule la rapidez tangencial.
Keith_Richards [23]

Answer:

v = 18.84 m/s

Explanation:

Let the question says,"The blade of a 0.30 m diameter fan rotates 1200 rev / min. calculate the tangential speed."

It is given that,

Diameter of a blade of fan is 0.3 m

Angular velocity of the fan is 1200 rev/min or 125.66 rad/s

We need to find the tangential speed of the fan. The relation is given by :

v=r\omega

r is radius of fan

v=0.15\times 125.66 \\\\v=18.84\ m/s

So, the tangential velocity is 18.84 m/s.

4 0
3 years ago
How many earth's are there in the universe??
loris [4]
Just one the earth is the only one like it it's completely unique
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3 years ago
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