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3 years ago

Velocity can be defined as ______ in a given direction

Physics

2 answers:

Ahat [919]3 years ago

6 0

Velocity can be defined as Speed in a given direction

solmaris [256]3 years ago

5 0

The correct answer is speed

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The wavelength of a light wave will affect the light’s

Tom [10]

Frequency and color.

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3 years ago

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AfilCa [17]

Mass of a sample of gas doesn't change, no matter what happens to its pressure, volume, or temperature.

7 0

3 years ago

An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons

vlada-n [284]

Answer:

$v=2.02\frac{m}{s}$

Explanation:

Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:

$K=U\\\frac{mv^2}{2}=\frac{kx^2}{2}$

Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:

$v=x\sqrt\frac{k}{m}\\v=2.63*10^{-2}m\sqrt{\frac{5.9*10^6\frac{N}{m}}{10^3kg}}\\v=2.02\frac{m}{s}$

8 0

3 years ago

he electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the

Paraphin [41]

Answer:

R₁ = 50.77 Ω

Explanation:

Since, we know that:

Electric Power = P = VI

but from Ohm's Law:

V = IR

(or) I = V/R

Therefore,

P = V²/R

(OR) R = V²/P

where,

V = Battery Voltage

R = Resistance of combination

FOR SERIES COMBINATION:

R = Rs = (57 V)²/48 W

Rs = 67.69 Ω

but, we know that:

Rs = R₁ + R₂

R₁ + R₂ = 67.69 Ω

R₁ = 67.69 Ω - R₂ __________ eqn (1)

FOR PARALLEL COMBINATION:

R = Rp = (57 V)²/256 W

Rp = 12.69 Ω

but, we know that:

Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω

using eqn (1) and value of R₁ + R₂, we get

Rp = 12.69 = R₂(67.69 - R₂)/67.69

859.08 = 67.69 R₂ - R₂²

R₂² - 67.69 R₂ + 859.08 = 0

Solving this quadratic equation we get the answers:

Either, R₂ = 50.76 Ω

Either, R₂ = 16.92 Ω

Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,

using this value in eqn (1), we get:

R₁ = 67.69 Ω - 16.92 Ω

4 0

3 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

4 years ago