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Roman55 [17]
2 years ago
14

Two spheres of equal mass, A and B, are projected off the edge of a 1.0 m bench.    Sphere A has a horizontal velocity of 10 m/s

and sphere B has a horizontal velocity of 5 m/s.
__ 5.    If both spheres leave the edge of the table at the same instant, sphere A will land
a.    at some time after sphere B.
b.    at the same time as sphere B.
c.    at some time before sphere B.
d.    There is not enough information to decide.

__ 6.    If both spheres leave the edge of the table at the same instant, sphere A hits the floor at the spot marked X.  Sphere B will hit the floor
a.    at some point between the edge of the table and X.
b.    at some point past X.
c.    at the same distance from the table as X.
d.    there is not enough information to decide.
​
Physics
1 answer:
suter [353]2 years ago
4 0

Answer:

c. because A will land first becuase its heavier

and D.

Explanation:

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kykrilka [37]

Answer:

The jp2003parker guy is extremely wrong

So he says that the size wont matter and a physical change should occur, but how would the size change without having a physical change occur.

Explanation:

7 0
3 years ago
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What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?
salantis [7]

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the  speed over the ground  is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

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brainly.com/question/22271063?referrer=searchResults

8 0
3 years ago
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An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0
3 years ago
4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive
zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

|

B1 at 20km/h

|

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0
3 years ago
At 1 AM, a handsome astronomy instructor has just completed a late night dinner with Jennifer Garner and observes a nearly full
oksano4ka [1.4K]

Answer:

The correct answer is: waxing gibbous, 3 days

Explanation:

Waning quarter moon: hair removal time and bangs cuts.

The growing quarter as a moment of growth, development and evolution. On the contrary, the waning moon is associated with a time of completion, debugging or liquidation of pending issues.

We must take advantage of the influence of the lunar cycle in our favor according to the action we are going to take. If you have trouble growing your hair, try to go to the hairdresser in a crescent moon: it will grow faster. It is no nonsense. Since I cut my bangs to the Cleopatra, the touch-ups last me for another 1-1.5 weeks. As I reviewed the bangs in a growing room, in just a couple of weeks I was returning to the hairdresser.

That affects hair removal. There are many people who take appointments to the beautician to shave by consulting the lunar calendar. The hair removal done as soon as the dwindling is the best because it lasts longer, lasts for another week until the next appointment.

6 0
3 years ago
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