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choli [55]
3 years ago
6

Football player A has a mass of 210 pounds and is running at a rate of 5.0mi/hr. He collides with player B. Player B has a mass

of 190 pounds and is running in the same direction at 3.0 mi/hr. Which of the following statements is true?
A.) The momentum before the collision is equal to the momentum after the collision.
B.) The momentum before the collision is greater than the momentum after the collision.
C.) The momentum before the collision is less than the momentum after the collision.
Physics
2 answers:
LuckyWell [14K]3 years ago
8 0
The answer is b.) the momentum before the collision is greater than the momentum after the collision
NikAS [45]3 years ago
8 0

Answer: A.) The momentum before the collision is equal to the momentum after the collision.

Explanation:

From the Law of conservation of momentum, in a closed system, the net momentum is conserved.

Be it elastic collision or inelastic collision, whenever there is no external force acting in a system, the net momentum is always conserved.

Thus, initial momentum before collision is equal to the final momentum after collision.

Correct option is A.

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
What is the angle of refraction when a ray of light passes frm ethanol to air
77julia77 [94]

Answer: Your answer is<u> 1.36.</u>

Hope this helps!

5 0
2 years ago
Why is it important for body cell to have twice and many chromosomes as sex cells?
prisoha [69]

Answer:

Why sex cells only need half the number of chromosomes compared to other cells in the body?  As gametes are produced, the number of chromosomes must be reduced by half. Why? The zygote must contain genetic information from the mother and from the father, so the gametes must contain half of the chromosomes found in normal body cells.

Explanation:

Brainliest Plzz

8 0
2 years ago
Read 2 more answers
A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

4 0
3 years ago
A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. Th
Illusion [34]

Answer:

Explanation:

The power of each of the speakers is 0.535 W. At a distance d intensity of sound can be found by the following formula

Intensity of sound =  Power / 4π d²

= .535 / 4 x 3.14 x (27.3/2)²

= 2.286 x 10⁻⁴ J m⁻² s⁻¹

Intensity of sound due to other source = 5.715 x 10⁻⁵J m⁻² s⁻¹

Total intensity = 2 x 2.286 x 10⁻⁴J m⁻² s⁻¹

= 4.57 x 10⁻⁴J m⁻² s⁻¹

b ) In this case, man is standing at distances 18.15 m and 9.15 m from the sources .

The total intensity of sound reaching him is as follows

0.535 / (4 π x18.15² ) + 0.535 /  (4 π x9.15² )

= 1.293 x 10⁻⁴ + 5.087 x 10⁻⁴

= 6.38 x 10⁻⁴J m⁻² s⁻¹

5 0
3 years ago
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