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Leto [7]
3 years ago
12

A box has a length of 18 cm, a height of 19 cm, and a width of 20 cm. What is its volume?

Physics
2 answers:
Alex_Xolod [135]3 years ago
8 0
Volume of any rectangular solid = (length)(width)(height)

         (18cm)(20cm)(19cm) = 6,840 cm³
OLEGan [10]3 years ago
3 0
To find the volume, multiply them all together: (height x width x length)

18 x 19 x 20 = 6840cm³
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The periodic table is arranged by the number of neutrons is that statement true
DanielleElmas [232]

i guess that would be false cause ...

the elements in periodic table are arranged by atomic number which is basically the number of protons

hope that helps :)

6 0
2 years ago
Read 2 more answers
A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.47 m. A child appl
snow_tiger [21]

Answer:

255.34 J

Explanation:

Given,

Weight of disk = 805 N

radius = 1.47 m

Force applied by the child = 49 N

time = 2.95 s

KE = ?

mass of the disk

M = \dfrac{W}{g}= \dfrac{805}{9.81} = 82.059\ Kg

Moment of inertia of the disk

I = \dfrac{1}{2}Mr^2

I = \dfrac{1}{2}\times 82.059\times 1.47^2 =88.66\ kgm^2

Torque on the child

\tau = F \times r = 49 \times 1.47 = 72.03 Nm

Angular acceleration

\alpha = \dfrac{\tau}{I}=\dfrac{72.03}{88.66} = 0.812\ rad/s^2

So, angular speed at t = 2.95 s

\omega = \alpha t = 0.812 \times 2.95 = 2.4\ rad/s

Now, KE of the merry go round

KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2}\times 88.66\times 2.4^2 = 255.34 J

Hence, the Kinetic energy of the merry go round = 255.34 J

8 0
3 years ago
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side
pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0
3 years ago
A dog jumps 0.80 m to catch a treat. The dog's displacement vector is shown below.
Brrunno [24]

Answer:D

Explanation:

It was right o khan academy

6 0
2 years ago
A car is travelling at 15 m/s on a horizontal road and stopped after 4 s. The coefficient of kinetic friction between the tires
givi [52]

Answer:

Fr,= umg

umg= ma

a= v/t

umg= mv/t

u= v/gt= 0.38

3 0
2 years ago
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