All categories

3 years ago

34.9x46x809 I need help again

Physics

2 answers:

monitta3 years ago

7 0

Answer:

1298768.6

Explanation:

34.9 x 46 = 1605.4

1605.4 x 809 = 1298768.6

Irina-Kira [14]3 years ago

4 0

Answer:

hope that helps uh..☺

You might be interested in

Thank you in advance

Musya8 [376]

$m1 = \frac{- v2m2 - m2v1}{v2 - v1}$

8 0

3 years ago

A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe

inessss [21]

Answer:

1.5min

Explanation:

To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

By definition centripetal acceleration is given by

$a_c = \frac{V^2}{r}$

Where,

V = Tangencial velocity

r = radius

With our values we know that

$a_c = \frac{V^2}{r}$

$\frac{V^2}{r} = \frac{1}{10}g$

Therefore solving to find V, we have:

$V = \sqrt{\frac{1}{10}g*r}$

$V = \sqrt{\frac{9.81*200}{10}}$

$V = 14m/s$

For definition we know that the Time to complete are revolution is given by

$t = \frac{Perimeter}{Speed}$

$t = \frac{2\pi R}{V}$

$t = \frac{2\pi * 200}{14}$

$t = 1.5min$

6 0

3 years ago

Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?

bagirrra123 [75]

Reduce friction because friction just makes things harder

5 0

3 years ago

Read 2 more answers

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a)

mafiozo [28]

(a) The work done by the bird to raise the snake on the branch is equal to the product between the weight of the snake and the height of the branch above the ground:

$W_1 = mg h$

where $m=75 g=0.075 kg$ is the mass of the snake and $h=2.5 m$ is the height of the branch with respect to the ground. Substituting numbers into the equation, we find

$W_1 = (0.075 kg)(9.81 m/s^2)(2.5 m)=1.84 J$

(b) The work the bird did to raise its own centre of mass from the ground to the branch is equal to the product between the bird's weight and the height of the branch:

$W_2 = mgh$

where $m=350 g=0.35 kg$ is the mass of the bird. Substituting numbers into the equation, we find

$W_2 = (0.35 kg)(9.81 m/s^2)(2.5 m)=8.58 J$

6 0

3 years ago