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3 years ago

What is a biological study?

Physics

2 answers:

lys-0071 [83]3 years ago

7 0

Answer:

the study of living organisms utilizing the scientific method.

Explanation:

Fed [463]3 years ago

4 0

The study of living organisms

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Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago

A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k

Veronika [31]

Answer:

$14.98\ \text{kg m/s}$

$45.26^{\circ}$

Explanation:

$P_1$ = Initial momentum of the pin = 13 kg m/s

$P_i$ = Initial momentum of the ball = 18 kg m/s

$P_2$ = Momentum of the ball after hit

$55^{\circ}$ = Angle ball makes with the horizontal after hitting the pin

$\theta$ = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

$P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}$

Momentum in the y direction

$P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}$

$(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}$

The pin's resultant velocity is $14.98\ \text{kg m/s}$

$P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}$

The pin's resultant direction is $45.26^{\circ}$ below the horizontal or to the right.

4 0

3 years ago

1. The horizontal and vertical components of a projectile's velocity are

Anni [7]

The horizontal and vertical components of a projectile's velocity are independent of each other.

Answer: Option C

<u>Explanation:</u>

The path of a projectile is determined by two components of motion. They are termed as horizontal and the vertical components. Since both components velocity are perpendicular to each other, so it can stated that they are independent of each other.

Even it can seen that when the horizontal components of velocity is constant, then there will be change in the vertical components of velocity leading to free fall projectile path.

And in the absence of gravity, there will be change in the horizontal components of velocity with zero vertical component of velocity. Thus, the horizontal and the vertical components of a projectile’s velocity are seemed to be independent of each other.

5 0

3 years ago

in a longitudinal wave, the motion of the disturbance is in what direction relative to the wave motion?

VARVARA [1.3K]

Answer:

Longitudinal waves have the same direction of vibration as their direction of travel. This means that the movement of the medium is in the same direction as the motion of the wave.

4 0

2 years ago

9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T

andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

4 0

3 years ago