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3 years ago

Is anyone good at physics who can help me??

Physics

2 answers:

Elenna [48]3 years ago

8 0

Answer: 7

Explanation:

I might be wrong but let me know if it is or not.

Viefleur [7K]3 years ago

4 0

Answer:

1.07 m/s

Explanation:

This is a simple speed distance time question.

first, we need to determine the distance between -11 and 4.

(think of a number line).

We add 4 to 11 using absolute value (i.e taking away negatives), and we get 15 m, our distance. We are attempting to find speed, and we have our time.

Speed is distance/time.

Now we have our equation, we plug in our variables.

speed (aka velocity)= 15/14.

speed= 1.07142857143

If we round to 2 place values, our speed is 1.07 m/s.

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Answer:

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Explanation:

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How do you find the atomic number of a atom?

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An object moves in a straight path. Its position x as a function of time t is presented by the equation x(t) = at – bt2+c, where

ankoles [38]

Answer:

The answer is below

Explanation:

Given that:

x(t) = at – bt2+c

a) x(t) = at – bt2+c

Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:

x(t) = 1.4t - 0.06t² + 50

At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s$

b) x(t) = 1.4t - 0.06t² + 50

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s$

c) x(t) = 1.4t - 0.06t² + 50

At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

The average velocity (v) is given as:

$v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s$

6 0

3 years ago

If your body has a density of 955 kg/m3, what fraction of you will be submerged when floating gently in:

wel

Answer: (a) 99.5℅ submerged

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4 years ago

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aleksklad [387]

I would think it would be the same if you are weighting the dish and the Ice cube at the same time anyway. Not to sure though I'm a beginner and math is complicated for me sometimes.

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