All categories

2 years ago

Whats does 14.600 oz = to in grams

Physics

1 answer:

kifflom [539]2 years ago

8 0

Given that 1 ounce equals 28.35 grams, the value of 14.600 ounces in grams is approximately 413.9g.

<h3>What is the value of 14.600 oz in grams?</h3>

Note that:

1 ounce = 28.35 grams

Given that;

Value of mass = 14.600 oz
Value of mass in grams

Since 1 ounce = 28.35 grams, 14.600 oz will be;

14.600 oz = ( 14.600 × 28.35) grams

14.600 oz = 413.9g

Given that 1 ounce equals 28.35 grams, the value of 14.600 ounces in grams is approximately 413.9g.

Learn more about conversion of ounces to gram here: brainly.com/question/1117493

#SPJ1

You might be interested in

The two-way table below gives the thousands of commuters in Massachusetts in 2015 by transportation method and one-way length of

forsale [732]

Answer:

wow uh what? hdfgdfgf

Explanation:?

7 0

3 years ago

Read 2 more answers

- Explain why an aircraft will tend to lose height

olganol [36]

Answer:

the planes wings are lifting at an angle to gravity so the plane isn't lifting as much against gravity when it banks. some of the wing lift is going into turning the plane. :) so it needs more lift to bank and stay up

Explanation:

8 0

3 years ago

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

5 0

3 years ago

Radar use reflected microwaves to detect objects and measure their... what?

Sloan [31]

Height and depth..... for sure....

7 0

3 years ago

A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fir

bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0

3 years ago