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3 years ago

What determines the path that an object in projectile motion follows

Physics

2 answers:

Juli2301 [7.4K]3 years ago

6 0

ANSWER:

C) Gravity & Inertia.

Choices

A) gravity only

B) inertia and air resistance

C) gravity and inertia

D) gravity and air resistance

Ilia_Sergeevich [38]3 years ago

3 0

Direction and magnitude it is also determined by the gravitational acceleration any impacts or interruptions are ignored.

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a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration

slega [8]

Answer:

The answer of this question is =1.258*10-4

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2 years ago

Which of these is a part of the biosphere? a. wind b. lakes c. bacteria d. glaciers

Aleks [24]

B. lakes is a part of the biosphere
Biosphere is defined as "the regions of the surface, atmosphere, and hydrosphere of the earth (or analogous parts of other planets) occupied by living organisms." and a lake is the only thing on the list that could be occupied by a living thing.

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3 years ago

Read 2 more answers

Ideally, rewards should be given immediately and frequently but

snow_lady [41]

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3 years ago

A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds

Gwar [14]

Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

$v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\$

Now replacing the values we have:

$a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]$

3 0

3 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago