A bridge spanning a river is 98.0 meters above the water.

Suppose you are asked to find 91,321,972×0.007004. Doing a quick estimate by rounding the numbers in scientific notation, what v

slamgirl [31]

Answer:

Product, $P=9.1321972\times 10^7\times 7.004\times 10^{-3}$

P = 639619.091888

Explanation:

In this case, we need to find the product of two numbers. These are as follows :

$N_1=91321972=9.13\times 10^7$

$N_2=0.007004=7.004\times 10^{-3}$

On doing calculation, we get the product of two numbers as :

$P=N_1\times N_2$

$P=9.1321972\times 10^7\times 7.004\times 10^{-3}$

P = 639619.091888

In scientific notation the product of two number is given by :

$P=63\times 10^4$

Hence, this is the required solution.

3 0

3 years ago

What is the constant snell’s law

Marat540 [252]

Snell's law is defined as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media”.

3 0

2 years ago

Which characteristic should a good scientific question have? A) It should lead to a hypothesis that is not testable. B) it shoul

yarga [219]

Answer:

B. IT should have a very broad focus with many variables.

4 0

3 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

3 years ago

A hydraulic lift consists of two pistons that connect to each other by an incompressible fluid. If one piston has an area of 0.1

hjlf

You have:

A1= 0.15 m²
A2=6.0 m²
F1=130.0 N

To solve this problem you must use the pressure formula, as below:

P1=P2
F1/A1=F2/A2

Then, you have:

m2=A2xF1/A1xg

g=9.81 m/s² (This is the acceleration of gravity)

When you substitute the values in m2=A2xF1/A1xg, you obtain:

m2=(0.60 m² x 130.0 N)/(0.15 m² x 9.81 m/s²)

The answer is:

m2=530 Kg

5 0

4 years ago