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3 years ago

What is the force that opposes the movement of an object through water??

1 answer:

4 0

The force that opposes the movement of an object through water is called drag. This is a type of frictional force. This force normally depends on the density and the viscosity of the fluid in question. The liquid which has more density and more viscosity or stickiness will produce a greater amount of drag force on an object than a fluid that is less dense and less viscous in nature. River water normally has less drag than that of sea water.

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What causes objects to move or stay still (support your details with claims and evidence)

maxonik [38]

Force moves the object but if the same anyone force is applied to both sides then it doesn’t move

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3 years ago

__________ enable humans to see and detect the color of different objects.

Llana [10]

The correct answer is C.

6 0

4 years ago

Read 2 more answers

Why are some substances dissolve both polar and nonpolar substances?

dlinn [17]

Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

please mark me as the brainliest :)

use your brain :)

8 0

3 years ago

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for

Monica [59]

Question

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g

Part (b) what is the angle of banking of the highway? Give your answer in degrees

Answer:

a. Equation of Tangent

tan(θ) = v²/rg

b. Angle of the banking highway

θ = 0.087°

Explanation:

Given

Radius of the curve = r = 330m

Acceleration of gravity = g = 9.8m/s²

Velocity = v = 8km/h = 8 * 1000/3600

v = 2.22 m/s

a . Write an equation for the tangent of the highway's angle of banking

The Angle is calculated by

tan(θ) = v²/rg

θ = tan-1(v²/rg)

Part (b) what is the angle of banking of the highway? Give your answer in degrees

θ = tan-1(v²/rg)

Substituting the values of v,g and r

θ = tan-1(2.22²/(330 * 9.8)

θ = tan-1(0.001523933209647)

θ = 0.087314873580116°

θ = 0.087°

3 0

3 years ago

A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is

Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

$N=mg+\dfrac{mv^2}{r}$

$N=m(g+\dfrac{v^2}{r})$

$N=60\times (9.81+\dfrac{(10)^2}{30})$

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0

3 years ago