Abigail Folgers’s dying words were ?

Can anyone pls help me out in dis i am struggling in dis!

Step2247 [10]

The answer is C because all the other choices would cause disaster like pollute the water if you just throw it in the sink and it would smell terrible if you put it in the garbage etc...

3 0

3 years ago

Will give correct answer brainliest 5 kg m/s 8kg m/s 80 kg m/s 200 kg m/s

o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0

3 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

3 years ago

A 1000 kg elevator is accelerated upward at a rate of 0.70 m/s2. What is the tension in the cable pulling the elevator upward wh

Masteriza [31]

Answer:

Tension is also known as Force...

and Force is mass× acceleration.

so....1000×0.70=700N

5 0

4 years ago

Read 2 more answers

Please help me with this one

katen-ka-za [31]

Both of the pictures show how one picture is I believe in fall and the other one is in the forest and maybe in spring or summer. Also, the two pictures show 2 different animals. And maybe hunting for 2 different foods for them to survive.

* Hopefully this helps:) Mark me the brainliest:)

5 0

3 years ago