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otez555 [7]
3 years ago
6

Matthew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the ba

ll’s journey will gravity be the greatest force acting on the ball?
Physics
2 answers:
MakcuM [25]3 years ago
7 0
At the very top of the peak and on the way down
user100 [1]3 years ago
7 0
Hello There!

I'm assuming it is when it is at its highest point.

Hope This Helps You!
Good Luck :) 

- Hannah ❤
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How does the law of conservation of energy relate to the system shown in this model
egoroff_w [7]

Answer: The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another. Another way of stating this law of chemistry is to say the total energy of an isolated system.

Explanation:

3 0
3 years ago
Read 2 more answers
How energy and power are different ???
Naya [18.7K]

Answer:

Explanation:

Energy is what makes change happen and can be transferred form one object to another. ... Power is the rate at which energy is transferred. It is not energy but is often confused with energy. The watt is the most commonly used unit of measure for power.

6 0
2 years ago
A 1060-kg car moving west at 16 m/s collides with and locks onto a 1830-kg stationary car. determine the velocity of the cars ju
iren [92.7K]
M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore; 
(1060× 16) + (1830 ×0) = (1060 +1830) V
 16960 = 2890 V
      V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
5 0
3 years ago
Two small children decide it would be fun to toss a couple of large cats at each other. Cat A (7kg) is thrown at 7m/s and cat B
Alex777 [14]

Answer:

V=4.7m/s

Explanations:

Let Ma mass of cat A=7kg

Va velocity of cat A=7m/s

Mb mass of cat b=6.1kg

VB velocity of cat b=2m/s

From conservation of linear momentum

MaVa+MbVb=(Ma+Mb)V

7*7+6.1*2=(7+6.1)V

61.2=13.1V

V=4.7m/s

3 0
3 years ago
Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri
Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

         E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

 

Q = 24 pC = 24 10⁻¹² C

         E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

         E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

         E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

         E₂ = 7.2 N / C

let's find the difference between these two fields

         ΔE = E₂ -E₁

         ΔE = 7.2 - 5.4

         ΔE = 1.8 N / C

the minimum detection field is

         E_minimum = 0.77 N / C

        ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0
3 years ago
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