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2 years ago

Newtons second law in words??

Physics

1 answer:

Dennis_Churaev [7]2 years ago

5 0

Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object

Explanation:

it's newton's second law

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7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How

Lera25 [3.4K]

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

$Y = \frac{FL}{Ae}$

Part A

cross sectional area and applied force are the same as the original but the length is doubled

$\frac{FL_1}{A_1e_1} = \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} = \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o$

The second wire will be stretched by 2 mm

Part B

a third wire with the same length but twice the diameter of the first

$\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o$

e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm

3 0

4 years ago

Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy

Semenov [28]

Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

Q_lost = - Q_absorbed

The latent heat is

Q_absorbed = m L

The heat given by the body

Q_lost = M $c_{e}$ ΔT

where m is the mass of sweat and M is the mass of the body

m L = M c_{e} ΔT

m = M c_{e} ΔT / L

let's replace

m = 90 3.500 1.8 / 2.42 10⁶

m = 0.2343 kg

reduced to grams

m = 0.2342 kg (1000g / 1kg)

m = 23.42 g

the correct answer is c) 23 g

8 0

3 years ago

You work in a materials testing lab and your boss tells you to increase the temperature of a sample by 37.1 ∘c . the only thermo

Ivahew [28]

Each Celsius degree is the size of 1.8 Fahrenheit degrees. So you need dip your Fahrenheit thermometer into the sample, see where you're starting, and then warm it up to a temperature that reads (37.1 x 1.8) = 66.8 Fahreheit degrees higher.

7 0

3 years ago

In research in cardiology and exercise physiology, it is often important to know the mass of blood pumped by a person's heart in

frez [133]

Answer:

0.05081 kg

Explanation:

$m_1$ = Mass of blood