All categories

3 years ago

Please help me ! I have only 5 minutes left!

Physics

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

D: 4 m/s^2

Explanation:

(5, 10) (4, 6)

10 - 6 / 5 - 4 = 4/1 = 4

YESSIR I AM A LIFESAVER

You might be interested in

The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of

Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0

3 years ago

Question for the day!

Alex17521 [72]

Answer:

1. <--> A.

2. <--> C.

3. <--> D

4. <--> B.

explanation: i know my science!

3 0

3 years ago

Read 2 more answers

How much work is done if you push a box 200 meters with a force of 35 newtons answer?

VashaNatasha [74]

Work = force × distance
= 35 N × 200 m
= 7000 J

4 0

3 years ago

A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure

klasskru [66]

Answer:

$3x_0$

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

$d=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

$d=v_0 t = x_0$

in the second case, the horizontal velocity is increased to

$v_x' = 3v_0$

And so the new distance travelled will be

$d' = v_x' t = 3 v_0 t = 3 x_0$

So, the distance increases linearly with the horizontal velocity.

5 0

4 years ago

Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,

tamaranim1 [39]

Answer:

The ratio of $E_{app}$ and $E_{act}$ is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

$E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}$

$E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}$

Second equation is

$E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}$

We need to calculate the ratio of $E_{act}$ and $E_{app}$

Using formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}$

$\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}$

Put the value into the formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}$

$\dfrac{E_{app}}{E_{act}}=0.9754$

Hence, The ratio of $E_{app}$ and $E_{act}$ is 0.9754

8 0

3 years ago