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san4es73 [151]
3 years ago
11

Which term refers to the upper part of a glacier that receives the most snowfall?

Physics
1 answer:
liq [111]3 years ago
6 0
The answer is A) accumulation zone
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Bacteria vary somewhat in size, but a diameter of 2.9 μm is not unusual.
Sedbober [7]

Answer:

a) 6.4 x 10^-12 cm^3

b) 17 x 10^-6 mm^2

Explanation

a). The shape is assumed to be spherical The volume = volume of a sphere = \frac{4}{3} \pi r^3

3

4

πr

3

V = \frac{4}{3}*3.142* 1.15^3

3

4

∗3.142∗1.15

3

= 6.3715 μm^3

1 μm^3 = 10^-12 cm^3

6.3715 μm^3 = 6.3715 x 10^-12 cm^3

==> 6.4 x 10^-12 cm^3

8 0
2 years ago
Read 2 more answers
Help me and get points<br><br><br> What is the correct answer
Umnica [9.8K]
It’s the 3rd one......
8 0
3 years ago
0. The volume of a cube is 27cm3. Find the length of edge of cube.​
Dennis_Churaev [7]

Answer:

27 cm3

Explanation:

7 0
3 years ago
You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj
Oksana_A [137]

Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

6 0
3 years ago
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring t
dmitriy555 [2]

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

7 0
3 years ago
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