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Eva8 [605]
3 years ago
9

A ball rolls with a speed of 2.0m/s across a level table that is 1.0m above the floor. Upon reaching the edge of the table it fo

llows a parabolic path to the floor. How far along the floor is the landing spot from the table?
Physics
2 answers:
Mekhanik [1.2K]3 years ago
6 0

Answer:0.58 m

Explanation:

The initial velocity of the ball is u = 2.0 m/s

The height of the table is, h = 1.0 m

The ball falls in vertical direction under acceleration due to gravity.

Time taken for ball to hit the floor:

h= ut + 0.5gt² ( from the equation of motion)

1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²

Solving this for t,

t = 0.29 s ( we have neglected the negative value of t)

In the same time, the ball would cover a horizontal distance of :

s = u t

⇒s = 2.0 m/s×0.29 s = 0.58 m

Thus, the landing spot is 0.58 m away from the table.

sp2606 [1]3 years ago
5 0

Answer:

The distance covered by the ball is 0.9 m.

Explanation:

Given that,

Speed of ball = 2.0 m/s

Distance = 1.0 m

We need to calculate the time

Using equation of motion

h = ut+\dfrac{1}{2}gt^2

Here, u = initial velocity=0

t=\sqrt{\dfrac{3h}{g}}

Where, h = distance

g = acceleration due to gravity

t = time

Put the value into the formula

t=\sqrt{\dfrac{2\times1.0}{9.8}}

t=0.45\ sec

We need to calculate the distance of the ball

Using formula of velocity

d=v\times t

d=2.0\times0.45

d=0.9\ m

Hence, The distance covered by the ball is 0.9 m.

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Why do the waves have different speeds in different layers of Earth's surface?
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6 0
2 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

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3 years ago
Hii! help asap. i’ll give brainliest thanks!
oksian1 [2.3K]

Answer:

a i believe

Explanation:

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