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2 years ago

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1 answer:

Rus_ich [418]2 years ago

5 0

Answer:

8/9 parts

Explanation:

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State and briefly describe the applications of expansion<br>

zlopas [31]

When solid material expands in response to an increase in temperature (thermal expansion), it can increase in length in a process known as linear expansion. for an example application of expansion and contraction.

examples =
(1) Changing of shape and dimensions of objects such as doors.
(2) Wall collapsing due to bulging.
(3) Cracking of glass tumbler due to heating.
(4) Bursting of metal pipes carrying hot water or steam are some of the disadvantages of thermal expansion of matter.

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2 years ago

In the diagram, q1 and q2 are both +5.00*10^-8 C. What is the electric field at point P? Include a + or - sign to indicate the d

Damm [24]

Answer:

1389.1

Explanation:

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2 years ago

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

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3 years ago

A 100 kg cart is moving at 3 m/s. Calculate the cart’s kinetic energy.

STALIN [3.7K]

Answer:

450

Explanation:

Given,

Mass= 100kg

Velocity= 3 m/s

Kinetic Energy= ?

Kinetic Energy= 1/2 mv^2

= 1/2× 100× 3^2

= 1/2× 900

= 450.

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