What is one common product use microwaves dont have much timeee

Is it true or false that Science is a rigorous process, but is not subject to intense scrutiny by others

skelet666 [1.2K]

Answer:True

Explanation:

8 0

3 years ago

A syrup added to an empty container with a mass of 115.25g . When 0.100 pint of syrup is added, the total mass of the container

sesenic [268]

Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance. We calculate as follows:

Density = 182.48 g - 115.25 g / 0.100 pint ( 0.47 L / 1 pint ) = 67.23 g/0.047L
Density =1430.43 g/L

3 0

3 years ago

DedPeter [7]

Answer:

Both Electrical and Magnetic Forces take place between two charged objects

Explanation:

3 0

2 years ago

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

MrMuchimi

Answer: $37^{\circ}$ North of west

Explanation:

Given

40,000-ton luxury line traveling 20 knots towards west and

60,000 ton freighter traveling towards North with 10 knots

suppose v is the common velocity after collision

conserving momentum in west direction

$20\times 40,000=(20,000+60,000)v\cos \theta$

suppose the final velocity makes \theta angle with x axis

$v\cos \theta =10----1$

Conserving Momentum in North direction

$10\times 60,000=80,000\times v\sin\theta$

$v\sin \theta =\frac{15}{2}----2$

divide 1 and 2

$\tan \theta =\frac{3}{4}$

$\theta =37^{\circ}$

so search in the area $37^{\circ}$ North of west to find the ship

3 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$