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2 years ago

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A weightlifter lifts a set of 1250kg weights a vertical distance of 2m weight lifting contest. what potential energy do the weig

hts now possess

1 answer:

Juliette [100K]2 years ago

4 0

Answer:

24525 J

Explanation:

Using Mgh as the calculation for potential energy

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You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant spee

Lina20 [59]

Answer:

40.13491 m/s

Explanation:

$v_r$ = My speed = 35 m/s

v = Speed of sound in air = 343 Hz

$v_s$ = Speed of the police car

When the car is approaching

$f=f'\dfrac{v-v_r}{v-v_s}\\\Rightarrow 1340=f'\dfrac{343-35}{343-v_s}$

When the car is receding

$f=f'\dfrac{v+v_r}{v+v_s}\\\Rightarrow 1300=f'\dfrac{343+35}{343+v_s}$

Dividing the equations

$\dfrac{1340}{1300}=\dfrac{f'\dfrac{343-35}{343-v_s}}{f'\dfrac{343+35}{343+v_s}}\\\Rightarrow \dfrac{1340}{1300}=\dfrac{22\left(v_s+343\right)}{27\left(-v_s+343\right)}\\\Rightarrow -36180v_s+12409740-12409740=28600v_s+9809800-12409740\\\Rightarrow \frac{-64780v_s}{-64780}=\frac{-2599940}{-64780}\\\Rightarrow v_s=\frac{129997}{3239}\\\Rightarrow v_s=40.13491\ m/s$

The speed of the police car is 40.13491 m/s

5 0

2 years ago

A man runs at a velocity of 4.5 m/s for 15.0 min. When going up an increasingly steep hill, he slows down at a constant rate of

madreJ [45]

The man ran <u>4252.5 meters.</u>

Why?

To solve the problem, we need to divide the exercise into two movements, the first on while the was running at 4.5 m/s for 15 min, and then, while he was slowing down (going up because of the hill).

First movement: Running at 4.5 m/s for 15 min.

We need convert from minutes to seconds,

$1min=60seconds\\\\15min*\frac{60seconds}{1min}=900seconds$

Now, calculating the distance covered for the first movement, we have:

$x_{1}=0+v_{1}*t_{1}\\\\x_{1}=4.5\frac{m}{s}*900s=4050m$

So, we know that the man covered 4050m for the first movement, it will be our initial position for the second movement.

Second movement: acceleration -0.05m/s^2 (because he's slowing down) for 90 seconds, at 4.5m/s.

$x_{2}=x_{1}+v_{1}*t+\frac{1}{2}at^{2}\\\\x_{2}=4050m+4.5m\frac{m}{s}*90seconds-\frac{1}{2}*(0.05\frac{m}{s^{2}})*(90s)^{2}\\\\x_{2}=4050m+405m-(0.5*0.05\frac{m}{s^{2}}*8100s^{2})=4050m+405m-202.5m\\\\x_{2}=4252.5m$

Hence, we have that he ran 4252.5 m.

Have a nice day!

4 0

2 years ago

A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

Eratosthenes determined the circumference of Earth by conducting an experiment. Put his steps in order as they correlate to the

galben [10]

Solution: The correct order is: C, A, B

The statement of the problem:

How can we prove Earth is round and calculate its circumference?

Hypotheis:

If the sun casts shadows at different angles at the same time of day in different places, we can determine how much Earth curves.

If the Earth was flat, the angle measured at different places at the same time of the day would be same.

Observation:

In Syene, the sun's rays are vertical at noon. At the same time in Alexandria, the rays are 7.2 degrees from the vertical.

5 0

2 years ago