Ask question

Ask question

All categories

Sergio039 [100]

3 years ago

9

When an object is raised vertically, work is done against __________. What one word completes the sentence?

1 answer:

Rasek [7]3 years ago

7 0

Gravity Gravitational Force of Gravity

You might be interested in

Your friends sit in a sled in the snow. If you apply a force pf 75 N to them, they have an acceleration of 0.9 m/s ^ 2. What is

Elza [17]

Answer:

Mass of the sled in the snow 83.33 kg.

<u>Explanation</u>:

Given that,

Force applied to move the sled in the snow (F) = 75N

$\text { Acceleration }(a)=0.9 \mathrm{m} / \mathrm{s}^{2}$

We know that

Newton's second law of motion is

$\text { Force }=\text { mass } \times \text { acceleration }$

F = ma (Or "force" is equal to "mass" times "acceleration".)

So if we move this around we can isolate mass and get mass

$\text { Mass }=\frac{\text { force }}{\text { accelearation }}$

$\mathrm{M}=\frac{75}{0.9}$

M = 83.33 kg

Mass of the sled in the snow <u>83.33 kg.</u>

3 0

3 years ago

Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing of N 42º E from the western sta

olchik [2.2K]

Answer:

Explanation:

AB = 110 miles

Let the distance of the western station from fire is d.

As according to the diagram, use Sine law

$\frac{d}{Sin 15}=\frac{110}{Sin 133}$

d = 110 x 0.2588 / 0.73

d = 39 miles

8 0

3 years ago

When you step into a very full bath or hot tub, sometimes the water overflows. But when you step into a lake and swim, you don't

max2010maxim [7]

Answer:

This all simply due to the size of the two water bodies question and the your body size in a bath tub the water is small enough to allow the upthrust displace water out if the tub whereas in a lake the water is bigger and your body size is smaller to allow any noticeable upthrust that would cause an overflow

8 0

3 years ago

The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr

Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V= $A_1V_1+ A_2V_2+A_3V_3$

where suffix 1,2,3 refers to the three pipes.

=27L/min+16L/min+11 L/min

=54L/min

b) A x V=54L/min => $\frac{\pi }{4} d^2$ x v

d= 2 cm

$\frac{\pi }{4} d^2$ x v = 54

v= $\frac{4}{\pi }$ x $\frac{54}{2^2}$

-> $A_1$ x $V_1$ =27L/min => $\frac{\pi }{4} d_1^2$ x $v_1$

$d_1$ = 1.3cm

$\frac{\pi }{4} d^2$ x $v_1$ = 27

$v_1$ = $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$

Next is to find the ratio of speed i.e $\frac{v}{v_1}$

$\frac{4}{\pi }$ x $\frac{54}{2^2}$ / $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$ => $\frac{54}{27}$ $\frac{1.3^2}{2^2}$

$\frac{v}{v_1}$ = 0.845

8 0

3 years ago

Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

2 years ago