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3 years ago

What is Newton's second law

2 answers:

8 0

Newton's second law of motion pertains to the behavior of objects for which all existing forces are not<span> balanced. The second law states that the acceleration of an object is dependent upon two variables - the </span>net force<span> acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.</span>

astra-53 [7]3 years ago

6 0

Answer:

Newton's Second law of motion: "Force = mass × acceleration."

Explanation:

An example of the second law of motion is when you are riding a bike. Your bike is the mass and our leg muscles pushing on the pedals of your bike are the force.

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Read 2 more answers

Suppose a car of mass m is moving at a constant speed v of

SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

$\arcsin(v^2/(gR))$

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

$w_r = mg\sin(\theta)$

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

$F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g$

Also, the car is making a circular motion:

$a = \frac{v^2}{R}$

Combining the equations:

$\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}$

Finally the angle is

$\arcsin(v^2/(gR))$

4 0

3 years ago

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

3 years ago

Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi

zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

cos ( 360-30) = cos (-30) = x₁ / d

sin (-30) = y₁ / d

x₁ = d cos (-30)

y₁ = d sin (-30)

x₁ = 1.2 cos (-30) = 1,039 miles

y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

cos (270-20) = x₂ / d

cos (250) = y₂ / d

x₂ = 2.0 cos 250 = -0.684 miles

y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

cos (180 + 30) = x₃ / d

sin (210) = y₃ / d

x₃ = 1.6 cos 210 = -1.3856 miles

y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

cos (90 + 15) = x₄ / d

sin (105) = y₄ / d

x₄ = 2.6 cos 105 = -0.6729 miles

y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis (West-East direction)

X = x₁ + x₂ + x₃ + x₄

X = 1.039 -0.684 - 1.3846 - 0.6729

X = -1.7025 miles

Y = y₁ + y₂ + y₃ + y₄

Y = -0.6 -1.879 -0.8 +2.511

Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

D = √ (X² + Y²)

D = √ (1.7025² + 0.768²)

D = 1.8677 miles

let's use trigonometry to find the direction

tan θ = Y / X

θ = tan⁻¹ Y / x

θ = tan⁻¹ (0.768 / 1.7025)

θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

θ = 24.28º at South of West

4 0

3 years ago