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storchak [24]
3 years ago
13

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.

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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
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