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Cerrena [4.2K]
3 years ago
9

You have two balls of equal size and smoothness, and you can ignore air resistance. One is heavy, the other is much lighter. You

hold one in each hand at the same height above the ground. You release them at the same time. What will happen?
Physics
1 answer:
motikmotik3 years ago
4 0

Answer:

They will hit the ground at the same time.

Explanation:

By ignoring the opposing forces i.e. air resistant, both the heavy and light balls will fall with same acceleration due to gravity (g=9.8 m/s²) and g is independent of mass of the objects. Thus both will hit the ground at the same time.

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If a sample of a radioactive isotope has a half-life of 1 year, how much of the original sample will be left at the end of the s
Tasya [4]

Answer:

1/4 of the original

Explanation:

That would be TWO half lives:

1/2  * 1/2   = 1/4   <======= 1/4 would be left

4 0
9 months ago
Which statement is the best interpretation of the ray diagram shown below?
madreJ [45]

A) A concave mirror forming a larger, virtual image

Explanation:

The figure is missing; see attachment.

There are two types of mirror:

  • Concave (converging) mirrors: a concave mirror is a mirror that reflects the light inward
  • Convex (diverging) mirrors: a convex mirror is a mirror that reflects the light outward

The image formed by a mirror can also be of two types:

  • Real image: it is formed on the same side of the object, with respect to the mirror
  • Virtual image: it is formed on the opposite side of the object, with respect to the mirror

In the figure of this problem (see attachment), we see that:

- The mirror reflects the light from the object inward --> so it is a concave mirror

- The image is formed on the other side of the mirror --> it is a virtual image

So the correct option is

A) A concave mirror forming a larger, virtual image

Learn more about mirrors:

brainly.com/question/8737441

#LearnwithBrainly

7 0
3 years ago
Read 2 more answers
A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises
atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s²  * t²

-2.0 m = -4.9 m/s²  * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t      where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t²           y = 1.5 m       y0 = 0 m   t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

4 0
2 years ago
A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend
solong [7]

Answer:

When her hands extends, her momen of inertia is 4.28\ kg-m^2.

Explanation:

Given that,

Initial angular speed, \omega_i=8\ rad/s

Initial moment of inertia, I_1=100\ kg-m^2

Final angular speed, \omega_f=7\ rad/s

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

I_1\omega_1=I_2\omega_2

I_2 is final moment of inertia

I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2

So, when her hands extends, her momen of inertia is 4.28\ kg-m^2. Hence, this is the required solution.

7 0
3 years ago
An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm
horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

Q = \frac{\Delta P \pi r^4}{8\eta L}

now we know that

Q = 0.06 \times 10^{-6} m^3/s

radius = 0.2 mm

Length = 6.32 cm

\eta = 1\times 10^{-3} Pa s

now we have

6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}

3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}

\Delta P = 6028 Pa

now we have

P - 1500 = 6028 Pa

P = 7528 Pa

8 0
3 years ago
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