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1 year ago

A weightlifter lifts a 125-kg barbell straight up 1.15 m in 2.5 s. What was the power expended by the weightlifter?

Physics

1 answer:

labwork [276]1 year ago

7 0

Answer:

560

Explanation:

Round your answer to two significant figures.

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A child is playing with a spring toy, first stretching and then compressing it.

Reika [66]

Sorry, I’m only in 6th Grade, I don’t know the answer to this question.

5 0

1 year ago

A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe

galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m, y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}$

The velocity v is given by:

$\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}$

The acceleration a:

$\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}$

From the given values we get two equations:

$-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4$

We also know:

$\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}$

The magnitude of the acceleration a is:

$a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7$

The magnitude of position r is:

$r=R=68m$

Plugging in to the equation for a(t):

$\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}$

and solving for ω:

$|\omega|=0.66$

Now solve for time t:

$\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83$

Using the calculated values to compute v(t):

$\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}$

The speed of the car is:

$\sqrt{38.3^2 + (-23.5)^2} = 44.9$

The position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}$

5 0

1 year ago

Read 2 more answers

When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pull

Mandarinka [93]

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

$F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}$

The force of the sun is, therefore, $10^{10}$ times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

5 0

1 year ago

A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$