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Genrish500 [490]
3 years ago
10

Does your BMR increase decrease stay the same or dissapear as you age?

Physics
1 answer:
hammer [34]3 years ago
6 0

Answer:

it decreases because of the differences in hormones

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When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced?
Reika [66]

Answer: The correct answer is an emission line spectrum.

Explanation:

When the electrons are excited to the higher energy level, the energy is absorbed in this case.

When the electrons in the atom step down to lower energy levels in a thin cloud of hot gas then the radiation will emit.

The electron will lose energy in this case in the form of radiation. There will be an emission line spectrum.

8 0
3 years ago
Read 2 more answers
How to: speed distance time graph exam questions
romanna [79]

you would have to look up the formula and know what time and how fast and how far


7 0
3 years ago
Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot, isothermal metal substrate at Th = 300°C. The proper
Ad libitum [116K]

Answer:

74.52s

Explanation:

The solution is shown in the picture below

7 0
3 years ago
Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of th
sattari [20]

Answer:

48 m

Explanation:

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.

The eastbound train

First convert km/h to m/s

(97 × 1000)/3600

97000/3600

26.944444 m/s

As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion

V^2 = U^2 - 2as

O = 26.944^2 - 2 × 3.5 S

726 = 7S

S = 726/7

S1 = 103.7 m

The westbound train

Convert km/h to m/s

(127×1000)/3600

127000/3600

35.2778 m/s

Using third equation of motion

V^2 = U^2 - 2as

0 = 35.2778^2 - 2 × 4.2 × S

1244.52 = 8.4S

S = 1244.52/8.4

S2 = 148.2 m

S1 + S2 = 103.7 + 148.2 = 251.86

The distance between them once they stop will be

300 - 251.86 = 48.14 m

Therefore, the distance between them once they stop is 48 metres approximately.

5 0
3 years ago
A 4.00-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 27.0 N is require
NikAS [45]

Answer:

a)

135Nm⁻¹

b)

0.925 Hz

c)

1.2ms⁻¹

d)

0 m

e)

6.7ms⁻²

f)

\pm 0.2 m

Explanation:

a)

F = force required to hold the object at rest connected with stretched spring = 27 N

x = stretch in the spring from equilibrium position = 0.2 m

k = force constant of the spring

force required to hold the object at rest is same as the spring force , hence

F = k x

k = \frac{F}{x}

inserting the values

k = \frac{27}{0.2} = 135 Nm⁻¹

b)

frequency of the oscillations is given as

f =\frac{1}{2\pi }\sqrt{\frac{k}{m}}

inserting the values

f =\frac{1}{2(3.14) }\sqrt{\frac{135}{4}}\\f = 0.925 Hz

c)

A = Amplitude of oscillations = 0.2 m

w = angular frequency

Angular frequency is given as

w = 2\pi f = 2 (3.14) (0.925) = 5.8 rads⁻¹

Maximum speed of oscillation is given as

v_{max} = Aw

v_{max} = (0.2)(5.8)\\v_{max} = 1.2 ms⁻¹

d)

maximum speed of the object occurs at the equilibrium position, hence

x = 0 m

e)

Maximum acceleration of oscillation is given as

a_{max} = Aw^{2}

a_{max} = (0.2)(5.8)^{2}\\a_{max} = 6.7ms⁻²

f)

maximum acceleration occurs when the object is at extreme positions, hence

x = \pm 0.2 m

4 0
3 years ago
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