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3 years ago

15

A cannon Is pointed Upward 30 degrees and fires a cannonball with a speed of 100m/sec what is the component of the cannon balls

velocity parallel to the groud

1 answer:

Kamila [148]3 years ago

6 0

Answer:

86.6 m/s

Explanation:

vₓ = v cos θ

vₓ = 100 m/s cos 30°

vₓ = 86.6 m/s

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RideAnS [48]

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3 years ago

Read 2 more answers

Which action can be explained by physics?

steposvetlana [31]

Answer:

Actions that underlie mathematical rules, patterns or probability distributions.

For example how fast something falls at any given point or time.

More complex actions, such as human decision making in single individuals would be way too complicated to describe in physical terms.

But note that there can be physical models of such things as traffic when we can assume statistical knowledge of behavior.

Also physical models are used to plan such things as emergency exits in big stadiums, because many thousands of people can be described as particles flowing under a given pressure.

Every time we can gain good statistical knowledge and can therefore see patterns and rules in action, we can build theoretical models to make predictions and simulations (and games btw)

Since it's fair to say that mathematics is the science of patterns, it is plausible that physical descriptions often come in mathematical formulations, so that it can be understand as an efficient language of physics.

Neighboring disciplines like chemistry relay on physical theories to build on them,and then add shortcuts to fit their needs and interests, generating an own language for their field of study. But physicists may refer to them as anadd-on to physics.

Physics can basically explain all actions wich you can express in numbers.

But note that on a fundamental level physics describes 'how' things work, not necessarily 'why' they do it this way. The source of the basic and most fundamental physical constants and rules remains a mystery till this day.But of course there are theories on that as well, wich mostly can neither be proved or falsified.

The text is my own work and based of my general knowledge and quintessence of lectures on physics and other fields I attended.

(I would really appreciate the brainliest)

3 0

3 years ago

A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s

klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

$U = \frac{1}{2} Li^{2}$

The rate at which the energy is being stored in the inductor is given by

$\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1$

The current through the RL circuit is given by

$i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })$

Where τ is the the time constant and is given by

$\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16$

$i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}$

Therefore, eq. 1 becomes

$\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})$

At t = 0.13 seconds

$\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts$

(b) thermal energy is appearing in the resistance

The thermal energy is given by

$P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts$

(c) energy is being delivered by the battery?

The energy delivered by battery is

$P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts$

4 0

3 years ago

a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t

Oksi-84 [34.3K]

Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

L = 25 ( 1 + 12 x 10⁻⁶ Δt )

For brass ring

L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt

.004 = 7.9518 x 10⁻⁶ Δt

Δt = 4000 / 7.9518

= 503⁰C.

final temp = 503 + 15 = 518⁰C .

6 0

3 years ago

Oxygen and nutrients are moved through the body by the: a. circulatory system c. lymphatic system b. digestive system d. reprodu

vagabundo [1.1K]

A. The Circulatory System.

3 0

3 years ago