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2 years ago

Where is the centripetal force directed on a banked curve?

Physics

2 answers:

Alex777 [14]2 years ago

7 0

Answer:

The net force on a car traveling around a curve is the centripetal force, Fc = m v2 / r, directed toward the center of the curve.

Explanation:LET ME KNOW IF ITS WRONG. HAVE A NICE DAY!!!!!!!!!!

san4es73 [151]2 years ago

6 0

Answer: The net force on a car traveling around a curve is the centripetal force, Fc = m v2 / r, directed toward the center of the curve. The centripetal force is directed at right angles to the motion and also along the radius towards the centre of the circular path. In order for a car to move around a circular track, something must provide a centipetal force. One source of the centripetal force is friction between the tires and the road. This means that the tires do not need to supply any friction at all to keep the car from slipping sideways. ...

Hope this helps... stay safe and have a great day!!! :D

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8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-

Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{H}}$ (1)

Where:

$A=0.375 g$ is the final amount of the material

$A_{o}=3 g$ is the initial amount of the material

$t=1 h$ is the time elapsed

$H$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$0.375 g=(3 g)2^{\frac{-1h}{H}}$ (2)

$\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}})$ (4)

$-2.079=-\frac{1 h}{H}ln(2)$ (5)

Clearing $H$ :

$H=\frac{-1h}{-2.079}(0.693)$ (6)

Finally:

$h=0.333 h$ This is the half-life of the Bismuth-218 isotope

4 0

3 years ago

Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat

Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating (assume it goes upto h1 height )

using motion equations upwards

$s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m$

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s

2) motion under gravity (assume it goes upto h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

= 15.625 + 7.8125

= 23. 4375 m

3 0

3 years ago

A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive i

klio [65]

Answer:

1.08

Explanation:

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer , so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. = λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08

5 0

3 years ago

what is the dot product and cross product of of two vectors if the angle is between them is 90 degree?

DaniilM [7]

$\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}$