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Ede4ka [16]
3 years ago
12

From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

. Determine the uncertainty for the slope of the line.
Physics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

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A ball is thrown straight up with an initial velocity of 6.4 m/s. It travels for 0.64 seconds, and has a change of position of 2
Naya [18.7K]

Answer:

V = 0.9 m/s

Explanation:

The parameters given are:

Initial velocity U = 6.4 m/s

Time t = 0.64s

Height h = 2.05 m

To find the final velocity, let us use third equation of motion

V^2 = U^2 - 2gH

Since the ball is going upward, g will be negative

Substitute all the parameters into the formula

V^2 = 6.4^2 - 2 × 9.8 × 2.05

V^2 = 40.96 - 40.18

V^2 = 0.78

V = sqrt( 0.78)

V = 0.883 m/ s

V = 0.9 m/ s approximately

6 0
2 years ago
Which of the following is a likely life cycle of a star?
melomori [17]

Answer:

D.

Explanation:

But this just happen for big stars, like more than 20x the Sun mass.

Shortly: A nebula is a cloud of gas and dust, the material starts to be acummuleted and became a protostar (is like a big planet, almost  a star). With enought mass this is a star, burn hydrogen and transform it in Helium.

This occurs in Main Sequence, is about almost all the life time of a star. Then starts the lack of hydrogen. Gravity compress everything, pressure goes up and heat all. Too much energy, Helium get burned and the star grews fast, became a Red Giant. Time pass and the fuel is over, no more making fusion, gravity compress the star, too much strenght, colapses, neutron star.

If it have pretty mass, ok. If have more than like 2x Sun mass, became a blackhole.

7 0
3 years ago
How did Robert Whittaker change classification?
larisa [96]
He reorganized life into five kingdoms
7 0
2 years ago
Read 2 more answers
Suppose a shrimp has been put on the ground that has just been taken out of water.Now touch the shrimp from a distance by a stic
Alex777 [14]

Answer:

yes

Explanation:

I would say yes because it's exerting some type of force as it reacts to the stick to make it jump straight in an upwards direction

4 0
2 years ago
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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
liq [111]

Answer:

The speed of water must be expelled at 6.06 m/s

Explanation:

Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

p_f=p_i

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

p_{of}+p_{wf}=p_{oi}+p_{wi}

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

m_o*v_{of}+m_w*v_{wf}=m_o*v_{oi}+m_w*v_{wi}

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so v_{oi}=v_{wi} = 0\frac{m}{s}:

m_o*v_{of}+m_w*v_{wf}=0

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for v_{wf}:

v_{wf}=-\frac{m_o*v_{of}}{m_w}=-\frac{(6.00-1.85)*(2.70)}{1.85}

v_{wf}=-6.06\frac{m}{s}

The minus sign indicates the velocity of the water is opposite the velocity of the octopus.

3 0
2 years ago
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