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3 years ago

11

Your car gets a flat! You go from 90 kilometers per hour to a stop in 6 seconds. What is your rate of deceleration? (it's negati

ve!)
I need this ASAP!

1 answer:

defon3 years ago

8 0

Answer:

-15?

Explanation:

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To pop a balloon you stab it with a pencil. If the area of the pencil tip is .01 cm² and the pressure applied by the pencil to t

Romashka-Z-Leto [24]

Answer:

Push with force of 1N

Explanation:

I have explained in the paper.

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3 0

3 years ago

Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car i

Korvikt [17]

As per law of Heisenberg uncertainty law

product of uncertainty in position and uncertainty in momentum will be constant

$\Delta x . \Delta P = \frac{h}{4\pi}$

$\Delta x . m \Delta v = \frac{h}{4\pi}$

now plug in all data

$(5\times 0.3048). (2250 \times 0.454) \Delta v = \frac{6.6 \times 10^{-34}}{4\pi}$

$\Delta v = 3.37 \times 10^{-38} m/s$

So above is the uncertainty in velocity of the object

4 0

3 years ago

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A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°c and convection heat transfer

Elodia [21]

200 degrees because I need the points

4 0

3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

3 years ago