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3 years ago

Explain the interaction between centripetal force and inertia and what kind of motion this interaction causes.

Physics

2 answers:

quester [9]3 years ago

8 0

Answer:

Centripetal force pulls an object toward the center of a circle. Inertia is an object's tendency to keep moving in a straight line unless acted on by an outside force. Centripetal force causes an object to constantly change direction, so the combination of centripetal force and inertia causes an object to move in a circle.

Explanation:

EDGE

dalvyx [7]3 years ago

4 0

Answer:

Inertia is an object's tendency to keep moving in a straight line unless acted on by an outside force. Centripetal force causes an object to constantly change direction, so the combination of centripetal force and inertia causes an object to move in a circle. Hope it helps and your cute by the way

Explanation:

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What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s

Nikolay [14]

Answer: 20 m/s

Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:

Ekinetic=(1/2)*(m*v^2)

then E=0.5*30Kg*(20 m/s)^2=15*400=6000J

8 0

3 years ago

The type of friction between the pavement and the tires on a moving vehicle is called.._____friction.

FinnZ [79.3K]

Answer:

A) Kinetic

Explanation:

Kinetic friction is friction caused by motion. Kinetic energy is energy in motion.

5 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 cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G

Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time

acceleration(-g)= dv/dt

Solve it

dv = a dt

dv = -g dt

v - v₀ = -gt

v= dy/dt

dy = v dt

dy = ( v₀ - gt ) dt

y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

6 0

3 years ago

A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m

Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

$K.E+P.E=P.E+K.E$

$\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2$

$\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}$

Put the value into the formula

$\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}$

$\dfrac{v_{1}^2}{2}=18.62$

$v_{1}=\sqrt{2\times18.62}$

$v_{1}=6.10\ m/s$

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0

3 years ago