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Bumek [7]
2 years ago
14

A geologist is studying the shore along a river. She finds a pile of rocks at the base of a riverbank. These broken rock pieces

are the result of .
Physics
2 answers:
krok68 [10]2 years ago
4 0

erosion caused by the river flow

Sidana [21]2 years ago
4 0

A geologist is studying the shore along a river. She finds a pile of rocks at the base of a river bank. These broken rock pieces are the result of WEATHERING. A little farther along the river, the geologist find some smaller Pebbles and a very shallow, slow-moving section of the river. These rocks were moved to this location through the process of EROSION. The process of the Rocks actually being placed here it's called DEPOSITION

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A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the
UkoKoshka [18]

Answer:

 Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, KE = \frac{1}{2} mv^2, where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

  1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg

So, mass of bike = 38 kg.

3 0
2 years ago
A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled
sesenic [268]
Velocity =

(distance between start point and end point, regardless of the route traveled) / (time spent traveling).

That distance (called the "displacement"), is 10 meters, and almost exactly 1 hour is almost exactly 3,600 seconds. So the numerical value of the velocity during that time is

(10) / (3,600) = almost exactly 0.00278 m/s

= 2.78 x 10^-3 m/s.
3 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w
Artemon [7]

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

7 0
2 years ago
An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
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