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3 years ago

What are the SI units for liquid volume?

1 answer:

5 0

SI units for volume don't care whether you use them to describe
a solid, a liquid, a gas, or a volume of vacuum.

The units are the liter, and the cubic meter (1,000 liters).

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The answer to number 9 please

Delicious77 [7]

The magnitude is doubled. The direction doesn't change.

7 0

3 years ago

A friend insists that electric current is the same as electrical energy. Work with a partner to develop a rebuttal

kotykmax [81]

Answer: The electrons flowing through the wire are referred to as a quantity of electricity, and the flow of electricity is referred to as “an electric current.”

Explanation: Hope it Helps have a blessed day

4 0

3 years ago

A Transformer has a primary coil with 650 turns if the output voltage is 500 volts and an input voltage of 200 volts how many te

maxonik [38]

Answer:

1625

Explanation:

Ns/Np = Vs/Vp

Ns / 650 = 500/200

Ns = 2.5 x 650

Ns = 1625

6 0

3 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

$W=-Fd$

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

$F=\mu mg$

where

$\mu$ is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

$W=-\mu mg d$ (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

$-\frac{1}{2}mu^2 = -\mu mg d$

And solving for $\mu$ , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

$\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767$

7 0

3 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0

4 years ago