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3 years ago

If you have 4 kg of a sample with density of 1 897 g/ml what is the volume

1 answer:

3 0

We know, Volume = Mass / Density
Here, mass = 4 Kg = 4000 g
d = 1,897 g/ml

Substitute their values,
v = 4000 / 1897
v = 2.108 ml

In short, Your Answer would be 2.108 mL

Hope this helps!

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List how many significant figures the numbers contain. A. 205 cm. B. .00004 cm. C. 20 cm. D. 20. Cm.

stiv31 [10]

A. 205 cm
This has 3 significant figures: 2,0 and 5.

B. 0.00004 cm
This has 1 significant figure: 4

C. 20 cm
This has 2 significant figures: 2 and 0.

D. 20. cm
This has 2 significant figures: 2 and 0

7 0

3 years ago

What planet are people working on so they can move there

Nata [24]

Answer:

mars

Explanation:

just search it up

8 0

3 years ago

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A circuit has an overall resistance of 40 ohms and a current of 200mA.

svetoff [14.1K]

Answer:

Hey!

_________________

Voltage (V) = 0.8V

Current (I) = 200 mA = 200/10^3 = 2/10

Resistance = ?

Resistance = Voltage / Current

Voltage = Current × Resistance

0.8 = 2/10 × Resistance

0.8×10/2 = Resistance

8/2 = Resistance

Resistance = 4 ohm

_________________

Hope it helps...!!!

Explanation:

3 0

2 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers

A devout halloweener not only dressed as an astronaut, but travelled to the moon for the full experience. The astronaut jumps on

nikdorinn [45]

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)

x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

5 0

3 years ago