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stiks02 [169]
3 years ago
11

Why is accuracy important in dodgeball?

Physics
1 answer:
____ [38]3 years ago
8 0
Accuracy and power are the only things that matter in dodge ball. If you are not accurate, then your target does not get hit.
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An object is pulled with two forces, 10 N northward and 15 N southward. The direction of the net force is to the An object is pu
ValentinkaMS [17]

Answer:

check image

Explanation:

For any question related to newons law of motion first draw the free body diagram(FBD),

7 0
3 years ago
Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i
max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0
3 years ago
Which transition metals have more in common than other
jeka94

Answer:

A. alkali metals

Explanation:

im not sure about that

8 0
3 years ago
Read 2 more answers
A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o
Bond [772]

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

3 0
3 years ago
Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.
amm1812
-- Equations  #2  and  #6  are both the same equation,
and are both correct.

-- If you divide each side by  'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by  'frequency', you get Equation #3,
which is also correct. 
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.
7 0
3 years ago
Read 2 more answers
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