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Evgen [1.6K]
3 years ago
5

A car accelerates uniformly from rest and reaches a speed of 21.9 m/s in 8.99 s. Assume the diameter of a tire is 57.5 cm. (a) F

ind the number of revolutions the tire makes during this motion, assuming that no slipping occurs.
Physics
1 answer:
Marat540 [252]3 years ago
7 0

Answer:

Explanation:

initial velocity, u = 0 m/s

time, t = 8.99 s

final velocity, v = 21.9 m/s

diameter of tyre, D = 57.5 cm

(a) Let a is the acceleration, and n be the number f revolutions made by the tyre.

Use first equation of motion

v = u + at

21.9 = 0 + a x 8.99

a = 2.44 m/s²

Use third equation of motion

v² = u² + 2 a s

21.9 x 21.9 = 0 + 2 x 2.44 x s

s = 98.28 m

Circumference of wheel, S = π x D = 3.14 x 0.575 = 1.8055 m

number of revolutions, n = distance / circumference

n = 98.28 / 1.8055

n = 54.4

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Varvara68 [4.7K]

To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

W = mgh

Where,

m = mass

h = height

g = Gravitational constant

When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

h = 3h

Replacing in our equation we have to

W = mgh\\W = mg(3h)\\W = 3mgh

The correct answer is 4.

7 0
3 years ago
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N
ANTONII [103]

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

5 0
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Benny has 20 jellybeans and wants to share with his friends how many will each friend get? there are 5 friends.
trasher [3.6K]

Answer each friend will get 3.33333 repeating if he is included. if only his friends are getting them then each one gets 4

Explanation:

devide 20/6 and 20/5 respectively.

7 0
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2. An egg rolls off the kitchen counter with a velocity of 4m/s. If
-Dominant- [34]

Answer:

800m/s

Explanation:

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