What is the standard Hz rate (cycles per second) in the US?<br> 20 Hz<br> 80 Hz<br> 60 Hz<br> 50 Hz

Inertia increases as an object's _________ increases.

sammy [17]

Answer:

A. Mass

Explanation:

Inertia of an object is the resistance of the object to any change in its state of motion: it means that if an object is at rest, it tends to stay at rest for inertia (unless a net force acts on it), and if it is moving, it tends to continue moving with the same velocity, for inertia.

The inertia also describes how difficult it is to stop/accelerate an object, and it is directly proportional to the mass of the object: in fact, the larger the mass of an object, the more difficult it is to change its state of motion, and this means it has greater inertia.

6 0

3 years ago

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A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must i

kiruha [24]

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

7 0

3 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

bogdanovich [222]

Answer:

magnitude of force on charge 2Q = $\frac{KQ^{2} }{I^{2} }$

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force

║F║ = $\sqrt{f_{x}^{2} + f_{y}^{2} }$ = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = $\frac{KQ^{2} }{I^{2} }$

The direction of the force on charge 2Q is calculated as

tan ∅ = $\frac{f_{y} }{f_{x} }$ = 1.8284

therefore ∅ = $tan^{-1}$ 1.8284

= 61⁰

3 0

3 years ago

Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum

marta [7]

Answer:

$v_{top} = 61.96 \frac{mi}{h}$

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

$v_{top} = 27.7 \frac{m}{s}$

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

$1609.34 \ m = 1 \ mi$

Now, we can divide by 1609.34 meters on both sides:

$\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}$

The left sides equals 1, so

$1 = \frac{1 \ mi}{ 1609.34 \ m}$

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

$v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}$