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3 years ago

10

Which expression can be used to calculate centripetal acceleration?

1 answer:

meriva3 years ago

8 0

We can use multiple expressions to calculate centripetal acceleration, depending on the physical problem:

Centripetal acceleration with tangental velocity:

a=v²/r, where v is tangental velocity and r is the radius of the circular trajectory.

Centripetal acceleration with angular velocity:

a=ω²r, where ω is the angular velocity and r is the radius if the circular trajectory.

We can calcualte it from the centripetal force Fcp:

Fcp=m*a, so a=Fcp/m, where m is mass and a is the centripetal acceleration.

And also from the equation:

a=√(a₁²+a₂²), where a is total acceleration, a₁ is centripetal acceleration and a₂ is the tangental acceleration, so to get a₁, or centripetal acceleration:

a₁=√(a²-a₂²)

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Based on the data given, in what direction will the car accelerate?

skad [1K]

Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration.
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

(487-left + 632-right) - (632-right - 487-right) = 145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.

7 0

2 years ago

5. A man has a weight of 100 Newtons. How much work is done if he climbs 4 meters up a ladder? Plug numbers under the equation.

Virty [35]

Answer:

<h2>400 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 100 × 4

We have the final answer as

<h3>400 J</h3>

Hope this helps you

5 0

2 years ago

A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

ankoles [38]

Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J

4 0

2 years ago

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

ryzh [129]

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

$d_1 = v* t$

$d_1 = 18* 0.200 = 3.6 m$

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

$v_f^2 - v_i^2 = 2 a s$

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

$0^2 - 18^2 = 2*(-3.35)(s)$

$s = 48.35 m$

So total distance after which car will stop is

$d = d_1 + d_2$

$d = 48.35 + 3.6 = 51.95 m$

So car will not stop before the intersection as it is at distance 20 m

3 0

3 years ago

an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu

Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

$E1= k*q1/(h1-h2)..............(1)$

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1

$[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2$ \\

$E1=[tex]E= 359502+359502\\E=719004 V/m$ V/m[/tex]

similarly electric feild due to negative charge q2=-40

$[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2$ \\

$E2=359502V/m$

Total electric feild E at the aircraft is given as

$E= E1+ E2\\$ ...............(2)

Put values of E1 and E2 in equation2

$\\E=359502+359502\\E= 719004V/m\\$

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0

2 years ago