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3 years ago

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Find the horizontal and vertical components of a projectile that is fired with a velocity of 50m/s at 30 degrees relative to the

horizontal.

1 answer:

VashaNatasha [74]3 years ago

5 0

Find this answer on safari and if you can’t reply to me and i’ll help

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A scientist is studying a shock wave from an earthquake. What kind of wave is being studied? an electromagnetic wave a gamma ray

Anettt [7]

<span>So we want to know what kind of wave is the scientist is studying while studying earthquakes. Waves that produce earthquakes are mechanical waves. Gamma rays and radiowaves are both electromagnetic waves and don't require a medium but mechanical do. So the correct answer is mechanical waves.</span>

6 0

3 years ago

Read 2 more answers

A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an init

anygoal [31]

Answer:

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

$T=2\pi\sqrt{\frac{m}{K}}$

$T=2\pi\sqrt{\frac{0.3}{160}}$

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

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3 years ago

In a transformer, energy is carried from the primary coil to the secondary coil by:________

likoan [24]

In a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

To find the answer, we have to know more about the transformer.

<h3>How transformer works?</h3>

An item utilized in the transfer of electric energy is a transformer.
AC current is used for transmission.
It is frequently used to modify the supply voltage between circuits without altering the AC frequency.
The fundamentals of mutual and electromagnetic induction govern how the transformer operates.
Magnetic field through the primary coil changes when primary coil current varies. the iron core of the secondary coil likewise has a magnetic field.
EMF is therefore generated in the secondary coil.

Thus, we can conclude that, in a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

Learn more about the transformer here:

brainly.com/question/26787198

#SPJ4

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2 years ago

What is a learned behavior of a puppy

Ludmilka [50]

Its when the animal repeats what it is taught similar to a child you never really teach them to lie it just happens because they have seen other people do it

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3 years ago

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

Oxana [17]

Answer:

A) $I_{total}$ = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

I = $I_{cm}$ + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

$I_{total}$ = $I_{body}$ + 2 $I_{arm}$

$I_{body}$ = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

M = 7/8 m total

M = 7/8 64

M = 56 kg

The mass of the arms is

m’= 1/8 m total

m’= 1/8 64

m’= 8 kg

As it has two arms the mass of each arm is half

m = ½ m ’

m = 4 kg

The arms are very thin, we will approximate them as a particle

$I_{arm}$ = M D²

Let's write the equation

$I_{total}$ = ½ M R² + 2 (m D²)

Let's calculate

$I_{total}$ = ½ 56 0.20² + 2 4 0.20²

$I_{total}$ = 1.12 + 0.32

$I_{total}$ = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0

3 years ago