A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centripetal acceleration of the car? 391 mi/h2 40.0 mi/h2 5,000 mi/h2 48,800 mi/h2

Answers

Answer 1

Answer: $a_(centrip)= (v^2)/(R)= (125^2 )/(0.32)=48828 (mi)/(h^2)$
The last one is the closest

Answer 2

Answer:

48,800 mi/h2 is the right answer

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Could you please solve it with shiwing the full work1-

A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.

3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.

Answers

Answer:

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y

$v=\sqrt{v_(x) ^(2) +v_(y) ^(2) }$

$v=\sqrt{ 638.6^(2) +49 ^(2) }$

V= 640.48 m/s : total velocity in t= 5s

2. $v_(ox) =v_(o) cos33.2=20.9*cos33,2= 17.49 m/s$

$v_(oy)=v_(o)*sin33,2 =20.9*sin33,2=11.44 m/s$

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4. a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0 , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m

A radioactive nucleus has a half-life of 5*108 years. Assuming that a sample of rock (say, in an asteroid) solidified right after the solar system formed, approximately what fraction of the radioactive element should be left in the rock today? The age of the solar system is 4.5*109 years. (No calculator is necessary, but if it helps use it.)

Answers

Answer:

0.002

Explanation:

The half-life of the radioactive nucleus is related to its quantity, by the following equation:

$N_(t) = N_(0)2^{-t/t_(1/2)}$

Where:

N(t): is the quantity of the radioactive nucleus at time t

N₀: is the initial quantity of the radioactive nucleus

t: is the time = 4.5x10⁹ years

t(1/2): is the half-life of the radioactive nucleus = 5x10⁸ years

$(N_(t))/(N_(0)) = 2^{-4.5 \cdot 10^(9) y/5 \cdot 10^(8) y} = 2 \cdot 10^(-3)$

Therefore, the fraction of the radioactive element in the rock today is 0.002.

I hope it helps you!

One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

Learn more about energy of photons here:

brainly.com/question/34240307

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A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.

Answers

Answer:

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance = 100 cm

v = Image distance

f = Focal length = -23.9 cm (concave lens)

$h_u$ = Object height = 2.1 cm

Lens Equation

$(1)/(f)=(1)/(u)+(1)/(v)\n\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\n\Rightarrow (1)/(v)=(1)/(-23.9)-(1)/(100)\n\Rightarrow (1)/(v)=(-1239)/(23900) \n\Rightarrow v=(-23900)/(1239)=-19.29\ cm$

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

Magnification

$m=-(v)/(u)\n\Rightarrow m=-(-19.29)/(100)\n\Rightarrow m=0.1929$

$m=(h_v)/(h_u)\n\Rightarrow 0.1929=(h_v)/(2.1)\n\Rightarrow h_v=0.1929* 2.1=0.40509\ cm$

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?

Answers

Answer:

They had a well-developed agricultural system , they raised domesticated animals , they developed a writing system, and they used early forms of tools.

Final answer:

The claim that the ancient Egyptians were primitive and relied on aliens to build their monuments is inaccurate. Evidence from archaeology and history shows that the ancient Egyptians had advanced knowledge and skills in various fields. Their construction techniques and use of mathematics in building the pyramids are well-documented.

Explanation:

The claim that the ancient Egyptians were primitive and that their accomplishments, such as building the pyramids, were assisted by aliens is inaccurate. There is evidence from archaeology and history that ancient monuments were built by ancient people using their own ingenuity and capabilities. The ancient Egyptians had advanced knowledge in various fields including architecture, engineering, astronomy, and mathematics. For example, their construction techniques and use of mathematics in building the pyramids and other structures are well-documented.

Learn more about Ancient Egyptians

At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)

Answers

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

$R_t = R_o[1 + \alpha(T-T_o)]\n\n$

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

$R_t = R_o[1 + \alpha(T-T_o)]\n\n\R_t = 1.59*10^(-8)[1 + 0.0038(T-20)]$

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

$R_t,_(silver) = 2R_o,_(iron)\n\n1.59*10^(-8)[1 + 0.0038(T-20)] =(2 *9.71*10^(-8))\n\n\ \ (divide \ through \ by \ 1.59*10^(-8))\n\n1 + 0.0038(T-20) = 12.214\n\n1 + 0.0038T - 0.076 = 12.214\n\n0.0038T +0.924 = 12.214\n\n0.0038T = 12.214 - 0.924\n\n0.0038T = 11.29\n\nT = (11.29)/(0.0038) \n\nT = 2971.1 \ ^0C$

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

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