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3 years ago

12. Which of the following is the property of a system?

Physics

1 answer:

Sati [7]3 years ago

5 0

Answer:

Explanation:

Pressure, temperature are measurable properties and they are also known as physical properties.

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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

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3 years ago

What is the shape of a projectile

Cerrena [4.2K]

The answer I found was parabola?

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3 years ago

At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball

dedylja [7]

Answer:

Magnitude of net force will be 432.758 N

Explanation:

We have given x component of acceleration $a_x=760m/sec^2$

And vertical component of acceleration $a_y=770m/sec^2$

Mass of the ball m = 0.40 kg

So net acceleration $a=\sqrt{a_x^2+a_y^2}=\sqrt{760^2+770^2}=1081.896m/sec^2$

Now according to second law of motion

Force = mass × acceleration

So F = 0.40×1081.896 = 432.758 N

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3 years ago

Which instrument is used to measure liquid volume?

Alecsey [184]

Volumetric cylinders and volumetric flasks

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3 years ago

Read 2 more answers

As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0

Alika [10]

velocity of the physics instructor with respect to bus

$v = 6 m/s$

acceleration of the bus is given as

$a = 2 m/s^2$

acceleration of instructor with respect to bus is given as

$a = -2 m/s^2$

now the maximum distance that instructor will move with respect to bus is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 6^2 = 2(-2)(d)$

$-36 = - 4 d$

$d = 9 m$

so the position of the instructor with respect to door is exceed by

$\delta x = 9 - 6 = 3 m$

so it will be moved maximum by 3 m distance

7 0

3 years ago