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3 years ago

_____ is the frictional force needed to slow an object in motion

Physics

2 answers:

s2008m [1.1K]3 years ago

8 0

Answer:

<u>Drag force</u> is the frictional force needed to slow an object in motion

Explanation:

const2013 [10]3 years ago

8 0

Answer:

Kinetic friction

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Determine whether each of the following is exothermic or endothermic and indicate the sign of δh. A. Natural gas burning on a st

Oxana [17]

Answer:

A) exothermic, δh is negative

B) endothermic, δh is positive

C) endothermic, δh is positive

Explanation:

And endothermic process absorbs heat from its sorrounding, cooling the sorrounding down. Whereas an exothermic process releases heat to its sorrounding raising the temperature of the sorrounding system.

5 0

3 years ago

What is the Net Force?

marshall27 [118]

It is 800 N FN = 600N + 200 N = 800 N Answer to your question: The net force is all Newton's second law. It is the force that acts on a body or a particle. for example: It is the force we make when we push a car or something heavy that is in a straight line. .

3 0

3 years ago

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are inv

PolarNik [594]

Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The largest total area of the oil slick $A = 8.257 *10^{9} \ m^2$

Explanation:

From the question we are told that

The volume of oil the escaped is $V = 18000 \ L$

The refractive index of oil is $n_o = 1.1$

The refractive index of water is $n_w = 1.33$

The wavelength of the light is $\lambda = 485 \ nm = 485 * 10^{-9} \ m$

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

$d = m *\frac{\lambda}{2n_w}$

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n

It is usually take as 1 unless stated otherwise by the question

substituting value

$d = 1 * \frac{485 *10^{-9}}{2 * 1.1}$

$d = 218 nm$

The are can be mathematically evaluated as

$A = \frac{V}{d}$

Substituting values

$A = \frac{18000}{218*10^{-8}}$

$A = 8.257 *10^{9} \ m^2$

6 0

2 years ago

Calculate the pressure the fluid exerted on your diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is

erica [24]

Complete question:

A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.

Answer:

Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

Explanation:

Given;

density of water, ρ = 1000 kg/m³

diver's position below the surface of the water, h = 10 m

acceleration due to gravity, g = 9.8 m/s²

Let the atmospheric pressure, P₀ = 101325 Pa

The pressure 10 m below the surface of the water is calculated as;

P = P₀ + ρgh

P = 101325 Pa + (1000 x 9.8 x 10)Pa

P = 199325 Pa

P = 1.99 x 10⁵ Pa.

Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

5 0

2 years ago

A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr

natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

f ’= f₀ (v + v₀ / v- $v_{s}$ )

let's reduce the magnitude to the SI system

v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

f ‘= 100 (330+ 33.33 / 330-44.44)

f ’= 97.0 Hz

4 0

3 years ago