Pick the answer that would make the following statement true.

A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?

pishuonlain [190]

Answer: 800

Explanation:

1/2 x m x v^2 = m x g x h

KE = 10 x 10 x 8

KE= 800

3 0

2 years ago

What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?

eimsori [14]

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

4 0

3 years ago

8. A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation f

pychu [463]

(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

<h3>Initial velocity of the turtle</h3>

The initial velocity of the turtle is calculated as follows;

$v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s$

<h3>Initial acceleration of the turtle</h3>

The initial acceleration of the turtle is calculated as follows;

$a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2$

<h3>Initial position of the turtle</h3>

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

<h3>Time when the velocity becomes zero</h3>

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

<h3>Time taken to return to starting point</h3>

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

<h3>Time taken for the turtle to travel 30 cm</h3>

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

Learn more about velocity here: brainly.com/question/6504879

7 0

1 year ago

n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at

Volgvan

Answer:

a)-1.014x $10^{-7$ J

b)3.296 x $10^{-7$ J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

$U_{i$ = -GMaMb/ d

$U_{i$ = - 6.67 x $10^{-11}$ x 47 x 110/ 3.4 => -1.014x $10^{-7$ J

b) at d= 0.8m (3.4-2.6) and $U_{i$ =-1.014x $10^{-7$ J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

$K_{i$ + $U_{i$ = $K_{f$ + $U_{f$

As sphere starts from rest and sphere A is fixed at its place, therefore $K_{i$ is zero

$U_{i$ = $K_{f$ + $U_{f$

The final potential energy is

$U_{f$ = - GMaMb/d

Solving for ' $K_{f$ '

$K_{f$ = $U_{i$ + GMaMb/d => -1.014x $10^{-7$ + 6.67 x $10^{-11}$ x 47 x 110/ 0.8

$K_{f$ = 3.296 x $10^{-7$ J

6 0

2 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago