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amid [387]
3 years ago
12

The picture shows an orange falling from an orange tree

Physics
2 answers:
Alinara [238K]3 years ago
6 0

Answer:

A but not 100% certain.

Explanation:

you can eliminate C since it doesn't mention gravity, and B says gravity is pushing up on the orange. I think it is A because friction is going against the orange because it has to pass through the air, and that creates friction. I think that is the reason why things burn up when they enter the earths atmosphere.

il63 [147K]3 years ago
6 0

Answer:

Answer A is the correct one

Explanation:

The force of gravity is pulling down in the orange, while the force of resistance with the air is pushing up on the orange. This last force could be viewed as a type of "friction" force with the air.

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Plz help! ill give brainlyest to whoever gets this correct!
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If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?
joja [24]
Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: g=-9.81~m/s^2, and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
v_f^2 = v_i^2 + 2gS
where v_i=4~m/s is the initial vertical velocity of the athlete, v_f=0 is the vertical velocity of the athlete at the maximum height (and v_f=0~m/s at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
S= \frac{v_f^2-v_i^2}{2g}=0.82~m

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3 years ago
What makes a planet different from other celestial bodies?.
Serggg [28]

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Explanation:

The atmosphere, calculated to equations, are a lot more pulled down.

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3 years ago
Which form of energy body will possess when it is kept on the top of Hill
Oduvanchick [21]

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Read 2 more answers
PLEASEEEE HEEEEELP!!!!!
8090 [49]

Answer:

Before:

p_{truck}=16400\ kg.m/s

p_{car}=10000\ kg.m/s

After:

p_{truck}=8000\ kg.m/s

p_{car}=8400\ kg.m/s

v_{fcar}=8.4\ m/s

F=9333.33 \ Nw

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

p_1=m_1v_{1o}+m_2v_{2o}

After the collision, they have speeds of v1f and v2f and the total momentum is

p_2=m_1v_{1f}+m_2v_{2f}

Impulse J is defined as

J=F.t

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

J=\Delta p

As the total momentum is conserved:

p_1=p_2

m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}

We can compute the speed of the second object by solving the above equation for v2f

\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{  m_2 }

The given data is

m_1=2000\ kg

m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s

a) The impulse will be computed at the very end of the answer

b) Before the collision

p_{truck}=2000\cdot 8.2=16400\ kg.m/s

p_{car}=1000\cdot 0=0\ kg.m/s

c) After collision

p_{truck}=2000\cdot 4=8000\ kg.m/s

Compute the car's speed:

\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }

v_{2f}=8.4\ m/s

And the car's momentum is

p_{car}=1000\cdot 8.4=8400\ kg.m/s

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s

The force can be computed as

\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw

The force on the car has the same magnitude and opposite sign

7 0
4 years ago
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