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Artyom0805 [142]

3 years ago

11

Two charged objects have a repulsive force of 0.050 N. If the charge of both of the objects is doubled, then what is the new for

ce?

1 answer:

grandymaker [24]3 years ago

7 0

Answer: 4 times grease

Explanation: Force F= C · q1·q2/r². C = Coulomb's constant.

If charges double you have 2q1 and 2q2.

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The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the ele

Blizzard [7]

Answer:

The magnitude of the electric field intensity is $E = 7.89 *10^{6} \ V/m$

Explanation:

From the question we are told that

The voltage is $\epsilon = 72.7 \ mV = 72.7 *10^{-3} V$

The thickness of the membrane is $t = 9.22 \ nm = 9.22 *10^{-9} \ m$

Generally the electric field intensity is mathematically represented as

$E = \frac{\epsilon }{t}$

substituting values

$E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}$

$E = 7.89 *10^{6} \ V/m$

8 0

3 years ago

I need help plzzz!!!!!

Masteriza [31]

Answer:

I truly don't know

Explanation:

5 0

3 years ago

The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.

Cerrena [4.2K]

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g ( g=9.81 m/s² )

F=ma

= (1510)*( 0.75*9.81)

= 11109.825 N

4 0

3 years ago

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

5 0

3 years ago

A uniform solid disk rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (En

Maru [420]

Answer:

aCM = (2/3)*g*Sin θ

Explanation:

Consider a uniform solid disk having mass M, radius R and rotational inertia I about its center of mass, rolling without slipping down an inclined plane.

In order to get the linear acceleration of the object’s center of mass, aCM ,

down the incline, we analyze this as follows:

The force of gravity (W = Mg) acting straight down is resolved into components parallel and perpendicular to the incline.

Since the object rolls without slipping there is a force of friction (Ff) acting on the object, at it’s point of contact with the incline, in the direction up the incline.

Newton’s 2nd Law gives then for acceleration down the incline

∑Fx' = m*aCM ⇒ m*g*Sin θ - Ff = m*aCM

The force of friction also causes a torque around the center of mass

having lever arm R so we can also write

τ = R*Ff = I*α

Solving for the friction, Ff = I*α / R

This is used in the expression derived from the 2nd Law:

m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM

The objects angular acceleration is related to the linear acceleration of the edge that contacts the incline by

a = R*α

Since the object rolls without slipping this has the same magnitude as aCM so we have that

α = aCM / R

Using this in

m*g*Sin θ - (I*α / R) = m*g*Sin θ - (I*(aCM / R) / R) = m*aCM

⇒ aCM = (m*g*Sin θ*R²) / (I + m*R²)

if I = (1/2)*m*R² (for a uniform solid disk)

we get

aCM = (2/3)*g*Sin θ

6 0

3 years ago