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MariettaO [177]
3 years ago
10

The momentum of an object is determined to be 7.2 ×× 10-3 kg⋅m/skg⋅m/s. Express this quantity as provided or use any equivalent

unit. (Note: 1 kgkg = 1000 gg).
Physics
1 answer:
7nadin3 [17]3 years ago
8 0

Answer:

Momentum, p = 720 g-cm/s

Explanation:

The momentum of an object is determined to be,

p=7.2\times 10^{-3}\ kg-m/s

We need to express this quantity in any equivalent units. We know that the conversions are as follows :

1 kg = 1000 g

and 1 m = 100 cm

p=7.2\times 10^{-3}\ kg-m/s=7.2\times 10^{-3}\times (1000\ g)\times (100\ cm)/s

p = 720 g-cm/s

So, the momentum of an object in any equivalent unit is 720 g-cm/s. Hence, this is the required solution.

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3 years ago
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In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra
shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0
3 years ago
The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a
BlackZzzverrR [31]

Answer:

E = 8.5 * 10^6 V/m

Explanation:

In general we have the following relation between the Electric Field and the Elecric Potential:

\int\limits^2_1 {\vec{E} \cdot\vec{dl}} = V_{2} -V_{1}

Due to the vector nature of the electric filed, we can only know the mean Electric field E across the membrane, and take it out from the integral, that is:

E = (ΔV)/L

Where L is the thickness of the membrane and ΔV is the potential difference.

Therefore:

E = 8.53933*10^6 V/m

rounding to the first tenth:

E = 8.5 * 10^6 V/m

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What has been happening to the cosmic microwave background radiation since the Big Bang?​
kondor19780726 [428]

Answer:

Explanation:

Cosmologists refer to a "surface of last scattering" when the CMB photons last hit matter; after that, the universe was too big. So when we map the CMB, we are looking back in time to 380,000 years after the Big Bang, just after the universe was opaque to radiation. But the CMB was first found by accident.

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3 years ago
A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r
elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

V_1=26.8 L

P_1=744mm Hg

T_1=31.2^{\circ} C=31.2+273=304.2K

P_2=385mmHg

T_2=-14.8^{\circ}=273-14.8=258.2K

We know that

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{P_2T_1}

Substitute the values

V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}

V_2=43.96 L

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0
3 years ago
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