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ivann1987 [24]
3 years ago
11

A 5.0 kg rock is dropped from the top of a building. The speed of the rock after it has fallen for 2.2 seconds is?

Physics
1 answer:
PSYCHO15rus [73]3 years ago
5 0
5x2.2/3 Because that's how you calculate velocity.

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If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan
ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

h=ut+\frac{1}{2}gt^2

here u=0 and t=time taken

h=\frac{1}{2}gt^2

for t=1 s

h_1=\frac{1}{2}g

for t=2 s

h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g

distance traveled in 2 nd sec=2g-\frac{1}{2}g=\frac{3}{2}g

for t=3 s

h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g

distance traveled in 3 rd sec=\frac{9}{2}g-2g=\frac{5}{2}g

so we can see that distance traveled in each successive second is increasing

5 0
3 years ago
What is the volume of a cube with length = 3 centimeters, width = 3
suter [353]

Answer:

D. 27 cubic centimeters

Explanation:

l*w*h

3*3*3= 27

4 0
3 years ago
A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going
Katyanochek1 [597]
Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
After another half-life, 500 are left.
After another half-life, 250 are left.
After another half-life, 125 are left.

That was FOUR half-lifes.

X = 4 .
8 0
3 years ago
Read 2 more answers
In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (65 m wide) and t
vfiekz [6]

Answer:

His launching angle was 14.72°

Explanation:

Please, see the figure for a graphic representation of the problem.

In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.

The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:

v = (vx, vy)

where vx is the component of v in the horizontal and vy is the component of v in the vertical.

In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:

sin angle = v0y / v0  then: v0y = v0 * sin angle

In the same way for vx:

vx = v0 * cos angle

Using the equation for velocity in the x-axis we can find the equation for the horizontal position:

dx / dt = v0 * cos angle

dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)

x - x0 = v0 t cos angle

x = x0 + v0 t cos angle

For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:

dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)

vy -v0y = g t

vy = v0y + g t

vy = v0 * sin angle + g t

The position will be:

dy/dt = v0 * sin angle + g t

dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)

y = y0 + v0 t sin angle + 1/2 g t²

The displacement vector at a time "t" will be:

r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)

If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)

r = (x0 + v0 t cos angle, 0)

The module of this vector will be the the total displacement (65 m)

module of r = \sqrt{(x0 + v0* t* cos angle)^{2} }  

65 m = x0 + v0 t cos angle ( x0 = 0)

65 m / v0 cos angle = t

Then, using the equation for the position in the y-axis:

y = y0 + v0 t sin angle + 1/2 g t²

0 =  y0 + v0 t sin angle + 1/2 g t²

replacing t =  65 m / v0 cos angle and y0 = 0

0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²  

cancelating v0:

0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)

-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)  

using g = -9.8 m/s²

-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²

sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²

(using trigonometric identity: sin x cos x = sin (2x) / 2

sin (2* angle) /2 = 0.25

sin (2* angle) = 0.49

2 * angle = 29.44

<u>angle = 14.72°</u>

3 0
3 years ago
Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m
Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:  

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12        ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4        ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

3 0
3 years ago
Read 2 more answers
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