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Alchen [17]
3 years ago
8

Please help answer #14!!!!!!!

Physics
1 answer:
jeka943 years ago
7 0

When a heavy football player and a light one run into each other, does the lighter player really exert as much force on the heavy player s the heavy player exerts on the light one. Yes. The interaction between the two players, the force each exerts on the other have equal strength.

Hope this helps you!!!

Please mark this as the brainliest one!!!

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The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3
Sav [38]

Answer:

a has units of distance

b  has units of distance over time

c  has units of distance over time^2

d has units of distance over time^3

Explanation:

Since the expression for the distance is:

x = a+b\,t+c\,t^2+d\,t^3

then:

a has units of distance

b  has units of distance over time

c  has units of distance over time^2

d has units of distance over time^3

because we are supposed to be able to add all of the terms and get a distance.  So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.

6 0
3 years ago
Which statement best describes how the temperature of the oceans surface water varies
Setler79 [48]
<span>heat capacity→ water has a high heat capacity, and salt water has an even higher one, so the temperatures of the oceans remain within a small range because As the heat rises you get more evaporation which actually cools the ocean down (Specific Heat of Water; Heat needed to break down hydrogen bonds)</span>
8 0
3 years ago
A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0
Lubov Fominskaja [6]

Answer:

t = 7.5 s

Explanation:

The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,

s₁ = Vi t + (0.5)gt²

where,

s₁ = distance traveled by police car

Vi = Initial Velocity = 0 m/s

t = time taken

Therefore,

s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

s₁ = 4.9 t²

Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:

s₂ = Vt = V(t + 1)

where,

s₂ = distance traveled by car

V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s

Therefore,

s₂ = 27.78 t + 27.78

Now, we know that at the time of meeting:

s₁ = s₂

4.9 t² = 27.78 t + 27.78

4.9 t² - 270.78 t - 27.78 = 0

solving the equation and choosing the positive root:

t = 6.5 s

since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.

t = 6.5 s + 1 s

<u>t = 7.5 s</u>

6 0
3 years ago
During a soccer game, a player grabs and holds an opponent's shirt outside of the penalty box. After the foul is called, what ki
Mila [183]
D direct free kick
Hope this helps
8 0
2 years ago
Read 2 more answers
When a certain gas under a pressure of 4.90 106 pa at 20.0°c is allowed to expand to 3.00 times its original volume, its final p
Elina [12.6K]

As per question the initial states of the gases are given as

INITIAL STATE:                                          FINAL STATE:

p_{1} =4.90106 pa                       p_{2} =1.06106 pa

v_{1} =v[say]                                 v_{2} =3v[say]

T_{1} =20 degree celcius  =293 K              T_{2} =?

AS  per combined gas equation obtained from the combination of Boyle's law and Charles law [Basic ideal gas laws]

              \frac{p_{1} v_{1} }{T_{1} } =\frac{p_{2}v_{2}  }{T_{2} }

Hence T_{2} =\frac{p_{2} v_{2}T_{1}  }{p_{1} v_{1} }

                    =\frac{1.06106*3v*293}{4.90106*v}

                    =190.3 K [ANS]

                 

   

6 0
3 years ago
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