Ask question

Ask question

All categories

3 years ago

8

Find the volume of the rectangular prism 9 cm 10 cm

1 answer:

sergeinik [125]3 years ago

3 0

Answer:

V= 90h cm³ where h is the height of the rectangular prism.

Explanation:

The formula for volume of a rectangular prism is ;

V=l*w*h where;

V=volume in cm³

l= length of prism=10cm

w =width of the prism = 9 cm

Assume the height of the prism as h cm then the volume will be;

V= 10* 9*h

V= 90h cm³

when the value of height of the prism is given, substitute that value with h to get the actual volume of the prism.

You might be interested in

In the ____ sector; occupations range from maintenance/service technicians, light vehicle technicians, and heavy vehicle technic

Len [333]

Answer:

B

Explanation:

becuase this is the best optimal answer when it comes to this particular situation of choice

5 0

2 years ago

From what side of the vessel should you never anchor

Pavel [41]

Answer:

Never anchor from the stern as this can cause the boat to swamp.

Explanation:

Brainliest pls

6 0

2 years ago

Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch

Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 $\frac{ft^{3}}{sec}$

$D_{1}$ = 6 inch=0.5 ft

$D_{2}$ =2 inch=0.1667 ft

As we know that Q=AV

$A_{1}\times V_{1}=A_{2}\times V_{2}$

So $V_{2}=\frac{Q}{A_2}$

$V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}$

$V_{2$ =0.687 ft/sec

We know that Head loss due to sudden contraction

$h_{l}=K\frac{V_{2}^2}{2g}$

If nothing is given then take K=0.5

So head loss $h_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}$

=0.00366 ft

So head loss=0.00366 ft

4 0

3 years ago

An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates wa

Phantasy [73]

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)= $\frac{Work produced}{Heat supplied}$

and

Heat Supplied ${\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )$

${Q_s}=3.3+4.4=7.7KW$

η= $\frac{3.3}{7.7}$

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1- $\frac{T_ {sink}}{T_ {source}}$

η=1- $\frac{273}{1023}$

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

5 0

3 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

3 years ago