All categories

3 years ago

A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to

Physics

1 answer:

nordsb [41]3 years ago

4 0

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

$5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s$

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

You might be interested in

How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?

tatyana61 [14]

Missing part in the text of the problem:
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
$Q=m C_s \Delta T$
where
m=1.8 g is the mass of the water
$C_s = 4.18 J/(g K)$ is the specific heat capacity of the water
$\Delta T=2.5 K$ is the increase in temperature.

Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
$E_1 = hf$
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
$N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons$

4 0

3 years ago

A student throws a 0.22 kg rock horizontally at 20.0 m/s from 10.0 m above the ground. Find the initial kinetic energy of the ro

LekaFEV [45]

Answer:

44J

Explanation:

Given parameters:

Mass of rock = 0.22kg

Initial velocity = 20m/s

Distance moved = 10m

Unknown:

Initial kinetic energy of the rock = ?

Solution:

To solve this problem, we need to understand that kinetic energy is the energy due to the motion of a body.

It is mathematically expressed as;

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Kinetic energy = $\frac{1}{2}$ x 0.22 x 20² = 44J

6 0

3 years ago

What the motion of an object that has an acceleration of 0 m/s

Degger [83]

<span>Everything in the system is stable and therefore the objects motion is stable. That is to say it is not changing what it is already doing. As far as i know zero times zero is still zero. In that case then the motion must be constant or stable.</span>

8 0

3 years ago

Someone please help on this one?

lawyer [7]

Explanation:

the lights on Galaxy watch shifted of uniform amounts to what the rate in of the spectrum,regardless of the distance from Earth

5 0

3 years ago

chapter 2 linear motion problems a student launches an arrow upward with an unknown initial velocity. the arrow takes 2.3 second

anzhelika [568]

Answer:

$V_{0}= 22.5\frac{m}{s}$ and Ymax=25.8m

Explanation:

Velocity at any time is given by $V=V_{0}sin\theta -gt$ . but when the arrow is on the top its velocity is zero and if it is launched upward the angle is 90°, so.

$0=V_{0} sin90-gt$

$V_{0}=gt=9.8\frac{m}{s^{2}}.2.3s=22.5\frac{m}{s}$

At the maximun height, position is given by $Ymax=V_{0}sin\theta.t-\frac{1}{2}gt^{2}$ , replacing $Ymax=22.5\frac{m}{s}x2.3s-\frac{1}{2}x9.8x(2.3)^{2}=25.8\frac{m}{s}$

5 0

3 years ago