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jek_recluse [69]
3 years ago
10

A student performs an experiment in which a 4kg object travels across a horizontal surface with an initial speed velocity of 6m/

s and a 2kg object travels across a horizontal surface with an initial speed velocity of −3m/s. The objects travel toward each other and collide. Data collected from the experiment were used to create the velocity versus time graph shown that contains information about both objects before and after the collision. The positive direction is considered to be to the right. Which of the following statements are true regarding the data from the experiment?
A- Both objects always travel in the positive direction during the experiment.

B- Kinetic energy is conserved during the experiment.

C- The position of the center of mass of the system remains constant during the experiment.

D- The velocity of the center of mass of the two-object system remains constant during the experiment.
Physics
1 answer:
Snezhnost [94]3 years ago
5 0

Answer:

The velocity of the center of mass of the two-object system remains constant during the experiment.

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If two objects are moved closer together so that the distance between them is one-third the original distance, what is the gravi
r-ruslan [8.4K]
Answer:
nine times the original gravitational force
Explanation:
The rule of the gravitational force between two bodies is shown in the attached images.
The parameters in the rule are:
Fg : the gravitational force between the two bodies
G : universal gravitational constant
m1 and m2 : the masses of the two bodies
r : the distance between the two bodies

From the given rule, we can notice that:
The force of attraction between the two bodies is inversely proportional to the square of the distance between them.
This means that:
As the distance decreases to 1/3 its original value, the gravitational force would increase by a factor of 9. 

Hope this helps :)

3 0
3 years ago
Read 2 more answers
A motorist drives south at 20m/s for 3 min then turns west and travels 25 m
Over [174]
If the motorist travelled at 20 meters a second for three minutes, 3 minutes is the same as 180 seconds, you multiply 20 times 180, which equals 3600, add 25, and the answer is 3625.
4 0
3 years ago
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Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of ma
Alex777 [14]

Answer:

N₂ = 503.8 N

Explanation:

given,

mass of bottom block = 37 Kg

mass of middle block = 18 Kg

mass of the top block = 16 Kg

force acting on the top block = 170 N

force on the block at top

N₁ be the normal force from block at middle

now,

N₁ = 170 + m g

N₁ = 170 + 16 x 9.8

now, force on block at middle

N₂ be the normal force exerted by the bottom block

N₂ = N₁ + m₂ g

N₂ = 326.8 + 18 x 9.8

N₂ = 503.8 N

hence, normal force by bottom block is equal to N₂ = 503.8 N

7 0
4 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
4 years ago
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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
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