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2 years ago

What kind of motion is the motion of the “flying bully”? How much and what is

1 answer:

7 0

<span>The flying bully is a move used in the Superhero Movie "Hancock", it is not a real motion in our universe. However, the direction would be towards the target object and the acceleration would be maximal.</span>

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Can an element be a molecule

Serga [27]

an element can make a molecule. so technically yes.

6 0

2 years ago

Read 2 more answers

A metal block has a mass of 5 grams and a volume of 1 cm3. What is the density of a piece of the same identical metal but with a

Fynjy0 [20]

Answer:

ρ = Mass / Volume definition of density

ρ = 5 g / 1cm^3 = 5 g / cm^3

Since the other object is made of the same metal its density is the same:

ρ = 5 g/cm^3

6 0

1 year ago

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

ale4655 [162]

Answer: The riders are subjected to 11.5 revolutions per minute

Explanation: Please see the attachments below

3 0

2 years ago

A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits

Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

5 0

2 years ago

A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is

Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2$

The acceleration of the craft should be 1.02234 m/s²

$F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N$

Weight of the craft

$W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N$

Thrust

$F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N$

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

3 0

2 years ago