All categories

2 years ago

What is centripetal force?

1 answer:

8 0

The centripetal force is force acting on a body in circular motion. In circular motion, velocity is always on tangent and if we took 2 different positions on a circle, the change on velocity is a vector pointing in the middle of circle. In circular motion velocity is constant, and acceleration lies on radius of circle pointing to te middle. This acceleration is called centripetal acceleration, and the force is centripetal.

You might be interested in

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

5 0

2 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the

scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = $\frac{kQ_{1}Q_{2} }{r^{2} }$

F= $\frac{K(Q/2)(3Q/4)}{r^{2} }$

how $\frac{KQ^{2} }{r^{2} }$ = 0.42

Then

F = $\frac{0.42*3}{8}$

F = 0.1575 N

3 0

2 years ago

A box is hanging from two strings. String #1 pulls up and left, making an angle of 50° with the horizontal on the left, and stri

creativ13 [48]

Answer:

mb = 3.75 kg

Explanation:

System of forces in balance

ΣFx =0

ΣFy = 0

Forces acting on the box

T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left

T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.

Wb :Weightt of the box (vertical downward)

x-y T₁ and T₂ components

T₁x= T₁cos50°

T₁y= T₁sin50°

T₂x= 30*cos75° = 7.76 N

T₂y= 30*sin75° = 28.98 N

Calculation of the Wb

ΣFx = 0

T₂x-T₁x = 0

T₂x=T₁x

7.76 = T₁cos50°

T₁ = 7.76 /cos50° = 12.07 N

ΣFy = 0

T₂y+T₁y-Wb = 0

28.98 + 12.07(cos50°) = Wb

Wb = 36.74 N

Calculation of the mb ( mass of the box)

Wb = mb* g

g: acceleration due to gravity = 9.8 m/s²

mb = Wb/g

mb = 36.74 /9.8

mb = 3.75 kg

8 0

2 years ago

A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f

Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

$W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)$

As we have already said, (1) is equal to the final kinetic energy of the puck:
⇒ Kf = 470 mJ (2)

8 0

2 years ago