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Rina8888 [55]
2 years ago
5

INQUIRY LAB

Physics
1 answer:
taurus [48]2 years ago
7 0

Answer:

u breathe lol nice ok bro

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A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0
3 years ago
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

5 0
3 years ago
Where do living organisms obtain energy from cell process
Yanka [14]
Hope this helps !?!!!!!!!!!!!!

4 0
3 years ago
In figure 16-2 is the temperature of the material within the cylinder greatest during the intake stroke, compression stroke, pow
erastovalidia [21]
Technically, we have no way of knowing that without seeing Figure 16-2.
So the question should be reported for incomplete content.  But I'm
going to take a wild stab at it anyway.

There's so much discussion of 'cylinder' and 'strokes' in the question,
I have a hunch that it's talking about the guts of a 4-stroke internal
combustion gasoline engine. 

If I'm right, then the temperature of the material within the cylinder is
greatest right after the spark ignites it.  At that instant, the material burns, 
explodes, expands violently, and drives the piston down with its stiff shot
of pressure.

This is obviously happening because of the great, sudden increase in
temperature when the material ignites and explodes.

It hits the piston with pressure, which leads directly to the power stroke.  
 
5 0
3 years ago
A car is moving at 80km/h accelerates to pass a truck. Five seconds later the car is moving at 100km/h. What is the Acceleration
labwork [276]

Answer/Explanation:

Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculation of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. We can just use the kinematic equations. Fortunately, we are given these values. So, we calculate as follows:

acceleration = v - v0 / t

acceleration = (80 mph - 50 mph) ( 1 h / 3600) / 5 s

acceleration = 1.67 x 10^-3 m / s^2

8 0
3 years ago
Read 2 more answers
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