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PIT_PIT [208]
3 years ago
9

A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?

Physics
1 answer:
GarryVolchara [31]3 years ago
3 0
Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns
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Mass of water= 357g density water= 1.0g/cm3
aliya0001 [1]

m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3

8 0
3 years ago
A water droplet falling in the atmosphere is spherical. Assume that as the droplet passes through a cloud, it acquires mass at a
ArbitrLikvidat [17]

Answer:

it b

Explanation:

bc A water droplet falling in the atmosphere is spherical

4 0
2 years ago
A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

3 0
3 years ago
Read 2 more answers
An unknown galaxy has a large flattened core. Which of the following classifications would best fit this galaxy's description? I
Alja [10]

Answer:

Spiral

i know it is so dont say nun people

Explanation:

7 0
3 years ago
Read 2 more answers
Some students set up a circuit and decided to measure the voltage at different points around
Fiesta28 [93]

Answer:

 V₁ = 6 V ,  V₂ = V₃ = 3 V

Explanation:

To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.

* a serial circuit there is only one path for current

in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term

* Parallel circuit in this there are two or more paths for the current

in this circuit the voltage is constant and the east is divided between each branch

with these principles let's analyze the proposed circuit

The DC battery is in parallel with resistor R1 and the equivalent of the other branch,

as in a parallel circuit the voltage is constant

               V₁ = 6 V

in the other branch (23) it forms a series construction, where the current is constant

               6 = iR₂ + iR₃

as they indicate that each resistance has the same value

              6 = 2 iR

              V = V₂ = V₃ = 3 V

3 0
3 years ago
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