All categories

3 years ago

How much space is taken up by a package with a length of 45 cm, a width of 10 cm, and a height of 8 cm?

Physics

2 answers:

soldier1979 [14.2K]3 years ago

6 0

The answer is C. 3,600cm cubed

lapo4ka [179]3 years ago

3 0

3600 cubic cm is the answer

You might be interested in

Please helppppppp!!!!!!!!!!!!!!!!!!!!!!!!

kykrilka [37]

Answer:

I think the Bulb No. 2 will stop emitting light if the bulb No. 1 burns out.

3 0

3 years ago

What happens when stress builds up at faults?

Harlamova29_29 [7]

Answer:

The answer is D because when the faults move that is the tectonic plates moving. So earth quakes will be forming when the fault moves.

Explanation:

3 0

3 years ago

Read 2 more answers

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

How should you behave while performing laboratory experiments?

Brrunno [24]

Explanation:

Always behave responsibly in the laboratory. Do not run around or play practical jokes. Always check the safety data of any chemicals you are going to use. Never smell, taste or touch chemicals unless instructed to do so.

4 0

3 years ago

Read 2 more answers

Which image shows both potential and kinetic energy?

Natasha2012 [34]

Answer:

The one in the middle

Explanation: i listened to the other person and i got it wrong, this is the answer for edge2020 sience review on energy!!!!

trust me its the middle one!!!!!

And everyone if ur not sure, like 100% sure about an answer dont answer at all cuz for 1: ur taking up a spot for others to answer. for 2: you could make people wrong. And for 3: its annoying. And 4: it makes stuff like this happen!

<u>NOT ARGUEING IM JUST PUTTING MY THOUGHTS AND OPINIONS OUT THERE ;)</u><em> many thanks.</em>

4 0

3 years ago

Read 2 more answers