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3 years ago

At what distance between two objects would you have the greatest gravitational force

1 answer:

6 0

The gravitational force is inversely proportional to the
square of the distance between their centers. So the
force is greatest when the distance is zero.

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A skateboarder starts from rest and maintains a constant acceleration of 0.50 m/s² for 8.4 s. What is the rider's displacement d

gregori [183]

Answer:

4.2m/s

Explanation:

0.50x8.4=4.2m/s

3 0

2 years ago

Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10

8090 [49]

A. 10 rations = 1 deca-ration.

b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.

c. 10⁻⁵ phones = 1 micro-phones.

d. 10⁻⁹ goats = 1 nano-goats.

e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.

4 0

3 years ago

A truck with 28-in.-diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile

solong [7]

Answer:

Angular speed ω=3771.4 rad/min

Revolution=5921 rpm

Explanation:

Given data

$d=28in\\r=d/2=28/2=14in\\v=50mi/hr$

To find

Angular speed ω

Revolution per minute N

Solution

First we need to convert the speed of truck to inches per mile

1 mile=63360 inches

1 hour=60 minutes

$v=(50*\frac{63360}{60} )\\v=52800in/min$

Now to solve for angular speed ω by substituting the speed v and radius r in below equation

$w=\frac{v}{r}\\ w=\frac{52800in/min}{14in}\\ w=3771.4rad/min$

To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation

$N=\frac{w}{2\pi }\\ N=\frac{3771.4rad/min}{2\pi }\\ N=5921RPM$

3 0

3 years ago

50 POINTS!

zysi [14]

The object is moving with constant or uniform acceleration and in average speed

The object is de accelerating

The object deaccelerated and came to rest so fast.

The object moves slowly first then accelerated.

The object accelerated at first so fast then move with constant acceleration then again accelerated .

7 0

3 years ago

At a hot air balloon race, a person on the ground shots a ball of confetti from a cannon to start the race. One of the balloons

Sati [7]

Answer:

a) The balloon and the ball will meet after 1.43 s.

b) The ball will reach the balloon at 15.7 m above the ground.

Explanation:

The height of the confetti ball is given by the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The height of the ball is given by this equation:

y = y0 + v · t

Where v is the constant velocity.

When the ball and the ballon meet, both heights are equal. Let´s consider the ground as the origin of the frame of reference so that y0 = 0:

y balloon = y ball

y0 + v · t = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

11 m/s · t = 18 m/s · t -1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m/s · t

0 = -4.9 m/s² · t² + 7 m/s · t

0 = t( -4.9 m/s² · t + 7 m/s)

t = 0 and

0 = -4.9 m/s² · t + 7 m/s

-7 m/s / - 4.9 m/s² = t

t = 1.43 s

They will meet after 1.43 s

b) Now let´s calculate the height of the balloon after 1.43 s

y = v · t

y = 11 m/s · 1.43 s = 15.7 m

The ball will reach the balloon at 15.7 m above the ground.

7 0

3 years ago