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ANTONII [103]
3 years ago
10

Which objects would sink in honey, which has a density of 1.4 g/cm³? Check all that apply.

Physics
2 answers:
bazaltina [42]3 years ago
6 0

Answer:

any object that has density more than 1.4

Explanation:

The object that has density more than 1.4 is denser than the honey

kykrilka [37]3 years ago
5 0

Answer:

D.

Explanation:

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At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at
Agata [3.3K]
We have the following equation for height:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 The value of the acceleration is:
 a = -g = -9.8 m / s ^ 2
 For t = 0 we have:
 h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
 h (0) = h0
 h0 = 0 (reference system equal to zero when the ball is hit).
 For t = 5.8 we have:
 h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
 (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
 vo = (1/2) * (9.8) * (5.8)
 vo = 28.42
 Substituting values we have:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
 Rewriting:
 h (t) = -4.9 * t ^ 2 + 28.42 * t
 The maximum height occurs when:
 h '(t) = -9.8 * t + 28.42
 -9.8 * t + 28.42 = 0
 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 
The ball was at maximum elevation when:
 
t = 2.9 seconds.
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The process by which the gene in the nucleotide suddenly changes its position is called.
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Mutation is the process by which the gene in the nucleotide suddenly changes its position

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A mutation is a permanent alteration of the nucleotide sequence of an organism, virus or other genetic elements.

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8 0
1 year ago
At higher speeds, how would you compensate for the decrease in field of vision
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5 0
3 years ago
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Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr
Illusion [34]

Answer:

F_{sprinter}=110.4N

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

W=F_{total}.d

So

W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

So

F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N  

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3 years ago
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