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1 year ago

What is the temperature when a solid begins to liquefy

Physics

1 answer:

MrRa [10]1 year ago

6 0

Answer:

Explanation:

The temperature is at its Melting Point - temperature at which a solid begins to liquefy.

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Any kind of wave spreads out after passing through a small enough gap in a barrier. This phenomenon is known as _________. a) di

bulgar [2K]

Answer:

a) diffraction

Explanation:

Diffraction occurs when waves pass through small openings, around obstacles or sharp edges.When an opaque object is between the point of light and a screen, the border between the shaded and illuminated regions on the screen is not defined. A careful inspection of the scrubber shows that a small amount of light is diverted to the shaded region. The region outside the shadow contains bright and dark altered bands, where the intensity of the first band is brighter than the region of uniform illumination.

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1 year ago

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Write a hypothesis about how the force applied to a cart affects the acceleration of the cart. Use the "if . . . then . . .becau

Lesechka [4]

"If one increases the force on an object, its acceleration increases too because the push it feels is greater"

We have the 2nd law of Newton that relates the 3 concepts; F=m*a. We have that if the mass of an object increases (put weight in luggage), the accelearation decreases; in fact it is inversely proportional to the mass. Hence if the mass is doubled, acceleration is halved. Accelerations is proportional to force; if one doubles the force, the acceleration doubles too.

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In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the

Roman55 [17]

The city monitors the steady rise of CO from various sources annually. In the year "C: 2019" (rounded off to the nearest integer) will the CO level exceed the permissible limit.

If this isn't the answer, let me know and i'll figure out what it is. But I believe this is it. :)

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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,

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

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0

1 year ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$