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3 years ago

5 grams is thes than greater than or equal too 508mg

Physics

2 answers:

guapka [62]3 years ago

7 0

Answer:

yes 5 is greater

Explanation:

makkiz [27]3 years ago

4 0

Answer: 5 grams is greater than 508 milligrams. Even though the number '508' is larger than '5,' you need to look at the units.

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In 1985, Said Aouita set the world record for the 1500m race in a time of 3 minute 29.46 second ? What was his average speed ?

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3 years ago

During which phase of a four stroke engine are waste gases removed?

Fed [463]

At the phase of D. Exhaust Stroke, the waste gases are removed during a four stroke engine. It is one of the phases of the four stroke engine. The cylinder from the fuel ignited during the compression step and removed from the cylinder.

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3 years ago

A baseball (m=140g) traveling 32m/s moves a fielders

posledela

Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average force exerted by the ball on the glove. It is solved using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

F = mg

$\dfrac{1}{2}mv^2=Fh$

$F=\dfrac{mv^2}{2h}$

$F=\dfrac{0.14\times (32)^2}{2\times 0.25}$

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

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3 years ago

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

levacccp [35]

Explanation:

The given data is as follows.

Mass, m = 75 g

Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

v = velocity after the impact

Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

$75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

= 13500 J

Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

Therefore, energy lost will be calculated as follows.

$\Delta E$ = E E'

= (13500 - 20) J

= 13480 J

And, n = $\frac{\Delta E}{E}$

= $\frac{13480}{13500} \times 100$

= 99.85

= 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0

3 years ago

Read 2 more answers

As pressure increases, temperature must ______________ for water to remain in a gaseous state.

Katyanochek1 [597]

As the pressure goes up, the temperature also goes up, and vice-versa.
<span>Also same as before, initial and final volumes and temperatures under constant pressure can be calculated.</span>

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3 years ago