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2 years ago

What two properties show that the drink is a fluid

2 answers:

7 0

Two properties that show that the drink is a fluid/liquid include its retention of volume and its conformation to the shape of its container.

Natali [406]2 years ago

7 0

Explanation:

Fluids have certain characteristics. Two of them are:

1. They do not resist any permanent deformation. In other words they can't resist shear (unaligned) forces applied on them.

2. They have ability to flow because of which they can take shape of the vessel they are kept in.

The drink also have these characteristics thus we can call that fluid.

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Block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is p

Tomtit [17]

Answer:

Please find attached

Explanation:

6 0

2 years ago

A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=35^{2}- 2 \times 9.8 \times h$

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0

2 years ago

A mover pushes a 255 kg piano

faust18 [17]

Answer:

$0.495 ms^{-2}$

Explanation:

According to the newton's second law of motion we can apply F=ma hear

Force = mass * acceleration

(assume the piano is moving left side )

←F = ma

$F_(pull)+ F_(push)= M*a\\77.5 + 48.7 = 255 *a\\a = 0.495 ms^{-2}$

7 0

2 years ago

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$