All categories

2 years ago

Two negative charges that are both -0.3C push each other apart with a force of 19.2 N. How far apart are the two charges?

Physics

1 answer:

Paha777 [63]2 years ago

8 0

<span>Using Coulomb's law: k*(-0.3)*(-0.3)/(d^2)=19.2 D is the distance between the two negative charges</span>

You might be interested in

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0

2 years ago

If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?

Vinvika [58]

The acceleration is0.25m/s^2

4 0

1 year ago

Read 2 more answers

Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder

Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

7 0

1 year ago

A pump moves water horizontally at a rate of 0.02 m3/s. Upstream of the pump where the pipe diameter is 90 mm, the pressure is 1

victus00 [196]

Answer:

the efficiency of hydralic is 79.88%

Explanation:

convert mm to m

1mm = (1/1000)m

diameter of pipe upsteam

d₁= 90mm= 0.09m

diameter of pipe downsteam

d₂= 30mm = 0.03m

finding velocity of upsteam

recall Q=A₁V₁

V₁=Q/A₁

V₁=3.14m/s

velocity of downsteam

V₂= Q/A₂

V₂= 28.29m/s

mass flow rate

m= ρQ

ρ is the density of water

m = 1000× 0.02

m= 20kg/s

the efficiency of hydralic is 79.88%

6 0

1 year ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

1 year ago