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1 year ago

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A discus thrower (Fig. P4.27, page 97) accelerates a discus from rest to a speed of 25.0 m/s by whirling it through 1.25 rev. As

sume the discus moves on the arc of a circle 1.00m in radius.(c) Calculate the time interval required for the discus to accelerate from rest to 25.0 m/s

1 answer:

Aleksandr-060686 [28]1 year ago

6 0

The time interval needed for the disk to rev from leave to 25.0 m/s

= 0.57 s

<h3>What is the time interval?</h3>

A larger period of time can be split up into multiple shorter, equal-length segments.
These are referred to as time periods. Consider the scenario where you wished to gauge a car's speed over an hour-long trip. You could break up an hour into ten-minute segments.
In Hz, frequency is defined as the number of cycles per second. Simply divide 1 by the frequency to determine the time interval for a known frequency (e.g., a frequency of 100 Hz has a time interval of 1/(100 Hz) = 0.01 seconds; 500 Hz has a time interval of 1/(500Hz) = 0.002 seconds, etc.).

Given:

vo = 0

v = 25.4 m/s

r = 0.95 m

n = 1.21 rev

A.

r*ω = v

ω = v/r

= 25.4/.95

= 26.74 rad/s

B.

θ = 2πn

= 2π × (1.21)

= 7.6 m

Using equation of motion,

ωf^2 = ωi^2 + 2αθ

(26.74)^2 = 2 × α × (7.6)

α = 715.03/15.2

= 47.04 rad/s^2

C.

Using equation of motion,

θ = ωi*t + 1/2*α*t^2

θ = 0 + 1/2*α*t^2

7.6 = 47.04/2 × t^2

= sqrt(0.323)

= 0.57 s

Therefore the correct answer is 0.57 s

To learn more about time interval, refer to:

brainly.com/question/479532

#SPJ4

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The density of aluminum is 2.7 g/cm3. A metal sample has a mass of 52.0 grams and a volume of 17.1 cubic centimeters. Could the

Fudgin [204]

Answer:
__________________________________________________
No; the sample could not be aluminum;
since the density of aluminum, " 2.7 g/cm³ " , is NOT close enough to the density of the sample, " 3.04 g/cm³ " .
________________________________________________
Explanation:
________________________________________________
Density is expressed as "mass per unit volume" ;

in which:
"mass, "m", is expressed in units of "g" (grams); and:
"Volume, "V", is expressed in units of "cm³ " (such as in this problem); or in units of "mL" ;
__________________________________________________
{Note the exact conversion: " 1 cm³ = 1 mL " .}.
__________________________________________________
The formula for density: D = m/V ;

Given: The density of aluminum is: 2.7 g/cm³.

Given: A sample has a mass of 52.0 g ; and Volume of 17.1 cm³ ; could it be aluminum?
_________________________________________________________
Let us divide the mass of the sample by the volume of the sample;
by using the formula:
___________________________________________
D = m / V ;

and see if the value is at, or very close to "2.7 g/cm³ ".

If it is, then it could be aluminum.
____________________________________________________
The density for the sample:

D = (52.0 / 17.1) g/cm³ = 3.0409356725146199 g/cm³ ;
→round to "3 significant figures" ;
= 3.04 g/cm³ .
_______________________________________________
No; the sample could not be aluminum; since the density of aluminum,
"2.7 g/cm³ " is NOT close enough to the density of the sample,
"3.04 g/cm³ " .
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3 years ago

Laser transmissions travel at a higher speed than the radio waves true or false

uranmaximum [27]

<span>False. Laser transmissions and radio waves are both forms of electromagnetic radiation. Both travel at approximately 3.00 x 10^8 meters per second. This is the case because wavelength and frequency are directly proportional to this value. Although laser light has a shorter wavelength than radio waves, it will travel at the same speed because the frequency will be greater.</span>

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3 years ago

A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the

Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) $8.40A/m^2$

(c) $\rho =15.52\times 10^{-9}ohm-m$

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

$A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2$

Resistance $R=11.9mohm=11.9\times 10^{-3}ohm$

Potential difference V = 33.7 volt

(A) current is equal to

$i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A$

(B) Current density is equal to

$J=\frac{i}{A}$

$J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2$

(c) Resistance is equal to

$R=\frac{\rho l}{A}$

$11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}$

$\rho =15.52\times 10^{-9}ohm-m$

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3 years ago

1. A car travels at 55 km/h for 6.0 hours. How far does it travel?

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Answer:

330 km

55*6= 330 km. An easy formula for this is multiplying time with speed.

Explanation:

330 kilometros

55 * 6 = 330 km. Una fórmula fácil para esto es multiplicar el tiempo por la velocidad.

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3 years ago

Read 2 more answers

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

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3 years ago