Please Help...........................................................

In a lab,two pieces of cellphane tape are placed on a lab table. They are quickly ripped off the lab table. Which would you pred

galben [10]

Answer:

they would repel each other

Explanation:

they have gained the same charge

4 0

3 years ago

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

LuckyWell [14K]

The answer is b.) the momentum before the collision is greater than the momentum after the collision

8 0

3 years ago

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Can Someone please help me! <br><br> What is deposition

7nadin3 [17]

Deposition is the geological process in which sediments, soil and rocks are added to a landform or landmass

7 0

3 years ago

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A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total

Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

$v_2 = v - 1$

now we know that total time of the motion is 19 hr

so we will have

$t_1 = \frac{90}{v}$

$t_2 = \frac{90}{v - 1}$

so we will have

$t_1 + t_2 = 19 hr$

$\frac{90}{v} + \frac{90}{v-1} = 19$

$90(2v - 1) = 19(v^2 - v)$

$19 v^2 - 199 v + 90 = 0$

by solving above equation we have

$v = 10 mph$

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0

3 years ago

On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total

Stella [2.4K]

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³

the volume V = 20,000/(1.2×9.8) = 1700 m³

3 0

3 years ago