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DanielleElmas [232]

2 years ago

13

What would be the weight (in Newtons) of a person with a mass of 80 kg on Earth, where the acceleration due to gravity is approx

imately 9.8 m/s/s? *
please help asap

1 answer:

defon2 years ago

6 0

Answer:

ml tayo 1v1 durugin kita

Explanation:

ano takot ml 1v1

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Carlos is making phosphorous trichloride using the equation below. He adds 15 g of phosphorus.

Degger [83]

Given:

The balanced chemical reaction of the synthesis of phosphorus trichloride:

2P + 3Cl2 ===> 2PCl3

Initial amount of phosphorus = 15 grams

The amount of product produced from 15 grams of phosphorus:

15 grams / 31 g/mol * (2/2) = 66.46 grams PCl3

The amount of chlorine is 44.31 grams, nearest to 45 grams.

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3 years ago

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Physics problem: piston? The piston of a hydraulic automobile lift is 0.300 m in diameter. 1. What gauge pressure, in pascals, i

k0ka [10]

Pressure is given by:
P=F/A
force=1200*10=12000 N
Area=pi*r^2=3.14*0.15^2
=0.07065 m^2
Thus;
P=12000/0.7065
P=169,851.38 N/m^2
but
1 N/m^2=1 Pa
thus our pressure is:
169,851.38 Pa

3 0

3 years ago

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What’s the temperature 414 K in degrees Celsius?

Alborosie

Answer:

414K − 273.15 = 140.85°C

Explanation:

7 0

3 years ago

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Which is the amount of matter in an object? a. weight b.mass c.volume d.pounds

hodyreva [135]

Answer:

(1.) (A.) wight

Explanation:

8 0

3 years ago

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the

Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

$a = \frac{v^{2} }{r}$

Where $a$ is the centripetal acceleration

$v$ is the speed

and $r$ is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be $v_{2}$ , that is

$v_{2} = 2v$

and without changing the length of the string means radius $r$ is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be $a_{2}$ .

Then we can write that

$a_{2} = \frac{v_{2}^{2} }{r}$

From

$a = \frac{v^{2} }{r}$

$v = \sqrt{ar}$

Recall that

$v_{2} = 2v$

∴ $v_{2} = 2\sqrt{ar}$

Now, we will put the value of $v_{2}$ into

$a_{2} = \frac{v_{2}^{2} }{r}$

Then,

$a_{2} = \frac{(2\sqrt{ar}) ^{2} }{r}$

$a_{2} = \frac{4ar }{r}$

$a_{2} = 4a$

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0

3 years ago