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2 years ago

Name two types of mechanical weathering in NewBedford

Physics

1 answer:

mario62 [17]2 years ago

8 0

Answer:

Explanation:

Mechanical weathering types

Mechanical weathering is the breaking down of rocks into smaller pieces without changing the composition of the minerals in the rock. This can be divided into four basic types – abrasion, pressure release, thermal expansion and contraction, and crystal growth.

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4. Brandon throws a tennis ball vertically upward. The ball returns to the point of release after 4.0 s. What is the

vekshin1

Answer:

-39.2m/s

Explanation:

Using the equation of motion;

v = u + at

Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g

v = u - gt

Since g = 9.8m/s²

t = 4.0s

u = 0m/s

v = 0 + (-9.8)(4)

v = -39.2m/s

Hence the speed of the ball before release is -39.2m/s

6 0

2 years ago

John runs around a 126.5 m circular track 3.5 times in 4.17 minutes. What is his average speed?

sdas [7]

Average speed = distance traveled / time

average speed = (126.5 m * 3.5 laps) / (4.17 min)

= 106.2 m/min

6 0

3 years ago

An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

jenyasd209 [6]

Answer:

a) i = -9.63 cm , h ’= .0.24075 cm erect

b) i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

1 / f = (n-1) (1 / R₁ - 1 / R₂)

1 / f = (1.70 -1)) 1 / ∞ - 1/13)

1 / f = 0.0538

f = - 18.57 cm

Now we can use the constructor equation

1 / f = 1 / o + 1 / i

1 / i = 1 / f - 1 / o

1 / i = -1 / 18.57 -1/20

1 / i = -0.1038 cm

I = -9.63 cm

For the height of the

image let's use magnification

m = h '/ h = - i / o

h ’= -h i / o

h ’= - 0.5 (-9.63) / 20

h ’= .0.24075 cm

b) we invert the lens

The focal length is

1 / f = (1.70 -1) (1/13 - 1 / int)

1 / f = 0.0538

f = 18.57 cm

1 / i = 1 / f -1 / o

1 / I = 1 / 18.57 - 1/20

1 / I = 3.85 10-3

i = 259.74 cm

h ’= - 0.5 259.74 / 20

h ’= 6.4935 cm

7 0

2 years ago

A cylindrical metal specimen having an original diameter of 11.77 mm and gauge length of 46.1 mm is pulled in tension until frac

Marta_Voda [28]

Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area $= \frac{(A_1 - A_2)}{A_1 } * 100$

$A_1 =\pi r^2 = \pi * 5.88^2=108.8 mm2$

$A_2=\pi * 3.98^2= 49.97 mm2$

% reduction in area = $\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %$

b)percentage elongation $= \frac{(66 - 46.1)}{46.1} *100 = 43.16 %$

5 0

2 years ago

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = 0 m/s

Acceleration = -g

Substituting

v = u + at

0 = u sin θ - g t

$t=\frac{usin\theta }{g}$

This is the time of flight.

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}$

This is the range.

In this problem

u = 30 m/s

g = 9.81 m/s²

θ = 45° - For maximum range

Substituting

$s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m$

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0

3 years ago