Usually, when the particle size is decreased, what will happen to the rate of dissolving?

A go-kart accelerates at a rate

stira [4]

Answer:

if it is at a speed of 12.5 m / s it would take 2 minutes

Explanation:

4 0

4 years ago

The plates of a parallel plate capacitor are separated by d = 1.6 cm. The potential of the negative plate is 0 V, and the potent

yaroslaw [1]

Answer:

The electric field between the plates is 1875 V/m.

Explanation:

Given that,

Separated d=1.6 cm

Potential of negative plate = 0 V

Potential halfway between the plates = 15 V

We need to calculate the total potential

Using formula of potential

$V=2\times\text{Potential halfway between the plates}$

Put the value into the formula

$V=2\times15$

$V=30\ V$

We need to calculate the electric field between the plates

Using formula of electric field

$E=\dfrac{V}{d}$

$E=\dfrac{30}{1.6\times10^{-2}}$

$E=1875\ V/m$

Direction is negative as the field always points from positive to negative plate

Hence, The electric field between the plates is 1875 V/m.

5 0

3 years ago

Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic

Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by $\frac{1}{2}kx^{2}$ where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, $\frac{1}{2}k(x1)^{2}$

⇒ $\frac{1}{2}k(\frac{w}{k})^{2}$

⇒ $\frac{w^{2}}{2k}$

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, $\frac{1}{2}2k(x2)^{2}$

⇒ $\frac{1}{2}2k(\frac{w}{2k})^{2}$

⇒ $\frac{w^{2}}{4k}$

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is $\frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=$ 2:1

4 0

3 years ago

A 13,900 N car traveling at 40.0 km/h rounds a curve of radius 1.80 ✕ 102 m.

miss Akunina [59]

a) $0.68 m/s^2$

b) 964.5 N

c) 0.069

Explanation:

a)

When an object is moving in a circular motion, the direction of its velocity is changing - therefore, it has an acceleration towards the center of the circle, called centripetal acceleration.

The magnitude of the centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circle

For the car in this problem:

v = 40.0 km/h = 11.1 m/s is the speed

r = 180 m is the radius of the circle

Substituting, we find the acceleration:

$a=\frac{11.1^2}{180}=0.68 m/s^2$

b)

The centripetal force is the force that keeps the object along its circular motion. It also acts towards the center of the circle, and it is given by

$F=ma$

where

m is the mass of the object

a is the centripetal acceleration

Here the weight of the car is

$W=mg=13,900 N$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

So the mass is

$m=\frac{W}{g}=\frac{13,900}{9.8}=1418.4 kg$

Therefore, the centripetal force is

$F=(1418.4)(0.68)=964.5 N$

c)

In this case, the force of static friction between the tires and the road provides the required centripetal force to keep the car in circular motion. This force is given by:

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

Equating the frictional force to the centripetal force,

$\mu mg=ma$

So we get:

$\mu=\frac{a}{g}$

And substitutng:

$a=0.68 m/s^2$ (centripetal acceleration)

$g=9.8 m/s^2$

We find:

$\mu=\frac{0.68}{9.8}=0.069$

4 0

3 years ago

At the instant a traffic light turns green, a car starts from rest with a given constant acceleration of 0.5 m/s squared. Just a

lubasha [3.4K]

Answer:

d= 1024 m

Explanation:

Kinematics of the car

The car moves with uniformly accelerated movement we apply the following formula:

d= v₀t+ (1/2)*a*t² Formula (1)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

a: acceleration in m/s²

Data

v₀₁= 0 :initial speed of the car

a₁ =0.5 m/s² : acceleration of the car

Kinematics equation of the car

We replace data in the formula (1) :

d= 0+ (1/2)*0.5*t²

d= (0.25)*t² Equation (1)

Kinematics of the bus

The car moves with uniformly movement ( constant speed) we apply the following formulas:

d= v*t Formula (2)

Where:

d:displacement in meters (m)

t : time in seconds (s)

Data

v = 16 m/s

Kinematics equation of the bus

We replace data in the formula (2) :

d= 16*t Equation (2)

<em>Problem development</em>

when the car passes the bus the elapsed time (t) and the distance (d) is equal for both.

Equation (1) = Equation (2)=d

(0.25)*t² =16*t We divide both sides of the equation by t

(0.25)*t = 16

t= 16/(0.25)

t= 64 s

We replace t in the equation (2)

d= 16*t

d= 16*(64)

d= 1024 m

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5 0

3 years ago