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3 years ago

8

If you pick up and carry a piece of heavy furniture which type of friction do you have with the floor

1 answer:

Phoenix [80]3 years ago

7 0

Sliding Friction hope its right

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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Convert 35 centimeters to inches

SIZIF [17.4K]

Answer:

13.7795276 Inches

Explanation:

5 0

3 years ago

1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second.

Alla [95]

Answer:

a.Distance = 150 m

b. Displacement = 50 m

Time lapsed = 5 seconds

Explanation:

a. Distance is the change in the position of an object.

The distance covered by the car = 100 + 50

= 150 m

b. Since displacement is a vector quantity,

Displacement of the car = 100 - 50

= 50 m due east

c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.

Thus, the time elapsed = 4 + 1

= 5 seconds

4 0

3 years ago

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00

vivado [14]

Answer:

7.5 m

Explanation:

$v$ = initial speed of the ball = 8 m/s

$\theta$ = angle of launch = 40° deg

Consider the motion along the vertical direction :

$v_{oy}$ = initial velocity along vertical direction = $v Sin\theta$ = 8 Sin40 = 5.14 m/s

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

$t$ = time of travel

$y$ = vertical displacement = - 1 m

Using the kinematics equation

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

$- 1 = (5.14)t + (0.5)(- 9.8)t^{2$

$t$ = 1.22 sec

Consider the motion along the horizontal direction :

$v_{ox}$ = initial velocity along horizontal direction = $v Sin\theta$ = 8 Cos40 = 6.13 m/s

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

$t$ = time of travel = 1.22 sec

$x$ = horizontal displacement = ?

Using the kinematics equation

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

$x = (6.13)(1.22) + (0.5)(0)(1.22)^{2$

$x$ = 7.5 m

4 0

3 years ago

Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre

Tanya [424]

Answer:

59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

<u>1. 18 mph 327º</u>

Horizontal component: 18 mph × cos (327º) = 15.10 mph

Vertical component: 18 mph × sin (327º) = - 9.80 mph

Vector notation:

$15.10\hat i-9.80\hat j$

<u>2. 4 mph 60º</u>

Horizontal component: 4 mph × cos (60º) = 2.00 mph

Vertical component: 4 mph × sin (60º) = 3.46 mph

Vector notation:

$2.00\hat i+3.46\hat j$

<u>3. Addition:</u>

You add the corresponding components:

$15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j$

To find the magnitude use Pythagorean theorem:

$\sqrt{17.1^2+6.34^2}= 18.23$

<u>4. Direction:</u>

Use the tangent ratio:

$tan(\alpha )=opposite/adjacent=3.46/2.00=1.73$

Find the inverse:

arctan (1.73) ≈ 59.97º

5 0

3 years ago