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3 years ago

11

A 0.010 kg clay ball thrown right at 16 m/s hits a marble moving left at 8.0 m/s, and sticks to it. Both move left at 3.2 m/s af

ter the collision. What is the mass of the marble?

2 answers:

4vir4ik [10]3 years ago

7 0

Answer:

0.040

Explanation:

Anettt [7]3 years ago

6 0

.……… it will be 7.5m/s

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A pencil is rolled off a table of height 0.92 m. If it has horizontal speed pf 1.4 m/s, how long does it take the pencil to reac

Alenkasestr [34]

The distance an object falls, from rest, in gravity is

   D = (1/2) (G) (T²)

'T' is the number seconds it falls.

In this problem,

   0.92 meter = (1/2) (9.8) (T²)

Divide each side by 4.9 : 0.92 / 4.9 = T²

Take the square root
of each side: √(0.92/4.9) = T

      0.433 sec = T

The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor. BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.

The pencil is in the air for 0.433 second.
In that time, it travels
   (0.433s) x (1.4 m/s) = 0.606 meter

from the edge of the table.

3 0

3 years ago

A balloonist happens to drop his metal ballpoint pen from his balloon as he is taking notes on his flight. Because his pen has a

Lerok [7]

So this is dealing with the conservation of energy. So you set kinetic energy equal to potential energy, so it looks like this:

1/2mv^2=mgh. The m's cancel out, so it is 1/2v^2=gh.

To find out what the height h is, divide g on both sides, so...

h=0.5v^2/g. v=22m/s, g=9.81m/s^2, so h=(0.5)(22^2)/(9.81)=24.67m

5 0

3 years ago

Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz

asambeis [7]

By using the Plancks-Einstein equation, we can find the energy;
E = hf
where h is the plancks constant = 6.63 x 10⁻³⁴
f = frequency = 3.55 x 10¹⁷hz
E = (6.63 x 10⁻³⁴) x (3.55 x 10¹⁷)
E = 2.354 x 10⁻¹⁶J

3 0

3 years ago

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is $\tau = 34.3 \ N\cdot m$

Explanation:

From the question we are told that

The mass of the steel ball is $m = 3.0 \ kg$

The length of arm is $l = 70 \ cm = 0.7 \ m$

The mass of the arm is $m_a = 4.0 \ kg$

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

$r = \frac{l}{2}$

=> $r = \frac{ 0.7}{2}$

=> $r = 0.35 \ m$

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

$\tau = m_a * g * r + m * g * L$

=> $\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70$

=> $\tau = 34.3 \ N\cdot m$

5 0

3 years ago

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

3 years ago