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Anettt [7]
3 years ago
6

An acorn falls from a tree. Its velocity just before it hits the ground is 28.2 m/s, downward. (acceleration of gravity is 9.81m

/s2). what is the acorn velocity at 1.4 s?
Physics
1 answer:
ss7ja [257]3 years ago
7 0

Answer:

12.74 ms^-1 download

Explanation:

v=28.2, a=9.81

start from rest u=0

v=u+at=0+(9.81)t=28.2

t=2.875...

it reach 1.4 second before hitting the ground:

t=1.4, u=0, a=9.81

v=u+at=0+(9.81)(1.4)=12.74

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D. O2 is removed from the atmosphere, and CO2 is released

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3 years ago
A spatially challenged goldfish swims along the x-axis only. Its initial position is 7.8 m. After swimming back and forth a whil
Verdich [7]

Answer:

<em>The fish displacement was -3.4 m</em>

Explanation:

<u>Distance and Displacement</u>

A moving object constantly travels some distance at defined periods of time. The total moved distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is obtained independently of the direction the object moves.

The displacement only takes into consideration the initial and final points of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.

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d = final position - initial position

We know the fist started at 7.8 m from a given reference along the x-axis.

After some undisclosed movements, it ends up at the position 4.4 m. Thus, the displacement is:

d = 4.4 m - 7.8 m = -3.4 m

This means the fish ended up to the left of the position it started from.

The fish displacement was -3.4 m

3 0
3 years ago
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
Zanzabum

Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

Generally according to the law of momentum conservation we have that

      m_b *  u_b  +  m_t *  u_t  =  m_b *  v_b  +  m_t  *  v_t

=>   0.024  *  1200  +  320 *  0  =  0.024 *  910   +  320  *  v_t

=>    v_t  =  0.02175 \  m/s

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