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Greeley [361]
3 years ago
5

During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What m

inimum speed did he need at launch if he was traveling at 6.8 m/s at the top of the arc
Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

Explanation:

Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.

Now, according to the law of conservation of energy the formula used is as follows.

mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s

As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.

v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s

Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

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A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. W
kramer

Answer:

Explanation:

Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is

F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes

F_n-w=ma  where F_n is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for F_n, that is what we will solve this literal equation for:

F_n=ma+w .  m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:

w = mg so

w = 28(9.8) and

w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually

w = 270 N.

Filling in the elevator equation:

F_n=28(.50)+270 and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:

F_n=14+270 and adding, by the rules, requires that we round to the tens place to get, finally:

F_n=280N  So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?

4 0
3 years ago
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he
lana [24]

Answer:8.28 km

Explanation:  

Given

First it drifts 45^{\circ} 2.5 km

r_1=2.5cos45 i+2.5sin45 j

Secondly it drifts 60^{\circ} 4.70 km

r_{12}=4.7cos60 i-4.7sin60 j

After that it drifted along east direction 5.1 km

r_{23}=5.1 i

After that it drifts 55^{\circ} 7.2 km

r_{34}=-7.2cos55 i-7.2sin55 j

After that it drifts 5^{\circ} 2.8 km

r_{54}=-2.8cos5 i+2.8sin5 j

r_{5O}=\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}+\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}

r_{5O}=2.299\hat{i}-7.95\hat{j}

|r_{5O}|=8.28 km

for direction

tan\theta =\frac{7.95}{2.299}=3.4580

\theta =73.87^{\circ} south of east

7 0
3 years ago
Read 2 more answers
Sam is pulling a box up to the second story of his apartment via a string. The box weighs 16.5 kg and starts from rest on the gr
Katyanochek1 [597]

Answer:

Weight (mass) = 16.5 kg

velocity = 0 m/a

acceleration =2.6 m/s^2

displacement = 13.2m

now,

acceleration = velocity/ time

2.6 = 0 / t

t = o / 2.6

t = o

8 0
3 years ago
What type of galaxy is M82 based on its appearance in the visible-light view?
SSSSS [86.1K]
<h2>Answer: irregular</h2>

According to Hubble  galaxies are classified into elliptical, spiral and irregular.

 

It should be noted this classification is based only on the visual appearance of the galaxy, and does not take into account other aspects, such as the rate of star formation or the activity of the galactic nucleus.  

The classification is as follows:  

1. Elliptical galaxies: Their main characteristic is that the concentration of stars decreases from the nucleus, which is small and very bright, towards its edges. In addition, they contain a large population of old stars, usually little gas and dust, and some newly formed stars.  

2. Spiral galaxies: They have the shape of flattened disks containing some old stars and also a large population of young stars, enough gas and dust, and molecular clouds that are the birthplace of the stars.  

3. Irregular Galaxies:  Galaxies that do not have well-defined structure and symmetry.  

In this context, galaxy M82 does not match with the first two types of galaxies, because it has not a defined shape.

Therefore, M82 is an  irregular galaxy.

4 0
3 years ago
A block of mass 20 kg sits on a ramp with an angle of 31 degrees above the horizontal. Assuming no friction, how fast will the b
Artyom0805 [142]

Answer:

2.98 m/s^2

Explanation:

I have done this before and it was a question on my physics test

4 0
3 years ago
Read 2 more answers
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