What is the answer to 7Hm a dm

What particles bond together to

yanalaym [24]

Answer:

atoms bond together and forms molecules

5 0

2 years ago

Read 2 more answers

1. thomas jefferson proposed using the length (l) of a simple pendulum whose period (t) was exactly 2 seconds as the definition

Alecsey [184]

The length of a 2 sec pendulum is 1 m.

Given that, initial length of the simple pendulum L₁ = 1 m

Initial time period T₁ = 2 sec

We need to find the length of the pendulum whose time period is 2 sec

T₂ = 2 sec

L₂ = ?

We know that the time period of the simple pendulum is given by the formula,

T = 2π√(L/g)

From the above relation, we can write T ∝ √L

T₁ / T₂ = √(L₁/L₂)

Making L₂ from the above relation, we have,

L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m

Thus, the length of a 2 sec pendulum is 1 m.

To know more about time period:

brainly.com/question/17350379

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7 0

1 year ago

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

Length of the string, L = 100 cm

The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0

3 years ago

A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta

Alex

Answer:

The depth will be equal to 6141.96 m

Explanation:

pressure on the submarine $P_{sea}$ = 62 MPa = 62 x 10^6 Pa

we also know that $P_{sea}$ = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

$P_{sea}$ = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine $P_{g}$ = 101 kPa = 101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure $P_{sea}$ , we have

62 x 10^6 = 10094.49h

depth h = 6141.96 m

7 0

3 years ago

A car is traveling down the highway at speed 5 m/s when the driver slams on the brakes and skids to a stop in a distance 80 m. A

Murljashka [212]

The car is very incredibly fast

8 0

3 years ago