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3 years ago

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A bomb is dropped from a bomber traveling at the speed of 120 km / h, destroying a military objective located at a distance of 2

000 m. How high was the plane traveling?

1 answer:

schepotkina [342]3 years ago

7 0

Answer:

18 km

Explanation:

Convert km/h to m/s:

120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s

The time it takes the bomb to travel the 2000 meters is:

2000 m / (33.3 m/s) = 60 s

So it takes 60 seconds for the bomb to fall. The distance it fell is therefore:

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²

Δy = 18,000 m

Δy = 18 km

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

sweet-ann [11.9K]

Answer:

(a) h = 1.27 Re

(b) h = 1.86 Re

Explanation:

Let M is the mass of earth and Re is the radius of earth.

initial velocity of projection, v = 0.462 ve

where, ve is the escape velocity of an object on earth surface.

(a)

The value of escape velocity is

$v_{e}=\sqrt{\frac{2GM}{R_{e}}}$

So, $v=0.462\times \sqrt{\frac{2GM}{R_{e}}}$ .... (1)

By using conservation of energy

(Kinetic energy + potential energy ) at the surface of earth = Potential energy at the height h.

where, h is the maximum height upto which the projectile reach

K.E at surface P.E at surface = P.E at the top

$\frac{1}{2}mv^{2}-\frac{GMm}{R_{e}}=-\frac{GMm}{h}$

By equation (1), substituting the value of v

$\frac{1}{2}\times 0.462^{2}\times \frac{2GM}{R_{e}}-\frac{GM}{R_{e}}=-\frac{GM}{h}$

$\frac{1}{2}\times 0.462^{2}\times \frac{2}{R_{e}}-\frac{1}{R_{e}}=-\frac{1}{h}$

h = 1.27 Re

(b)

initial kinetic energy = 0.462 times the kinetic energy required to escape

$\frac{1}{2}mv^{2}=0.462\times \frac{1}{2}mv_{e}^{2}$

$\frac{1}{2}mv^{2}=0.462\times \frac{1}{2}m\times \frac{2GM}{R_{e}}$

So, again by using the conservation of energy

Kinetic energy at the surface + Potential energy at the surface = Potential energy at the top

$0.462\times \frac{1}{2}m\times \frac{2GM}{R_{e}}-\frac{GMm}{R_{e}}=-\frac{GMm}{h}$

$0.462\times \frac{1}{R_{e}}-\frac{1}{R_{e}}=-\frac{1}{h}$

h = 1.86 Re

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3 years ago

A rectangular plate has a length of (22.4 ± 0.2) cm and a width of (8.4 ± 0.1) cm. Calculate the area of the plate, including it

a_sh-v [17]

To answer this problem, we can use the lower and upper limits altogether of the dimensions given. In this case, the minimum area is the product of 22.2 cm and 8.3 cm that is equal to 184.26 cm2 while the maximum area is equal to 22.6 cm times 8.5 cm equal to 192. 1 cm2. Plainly, 22.4 cm times 8.4 cm is equal to 188.16 cm2. The area is equal then to 188.16 +- 3.9 cm2

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3 years ago

What are 2 important statements about physical properties?

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Physical properties do not change the chemical nature of matter.
They are things you can observe and touch.

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Find the speed of a wave that has a wavelength of 6.0 m and a frequency of 2.0 Hz. V=?

Rzqust [24]

Frequency, υ = 2 Hz

Wavelength, λ = 6 m

We know that speed of a wave, v = υ × λ

⇒ v = 2 × 6 = 12 ms⁻¹

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3 years ago

"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1

kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

$\texttt{ }$

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>

$\boxed{ \omega = \omega_o + \alpha t }$

$\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }$

$\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }$

$\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }$

<em>where :</em>

<em>ω = final angular speed ( rad/s )</em>

<em>ω₀ = initial angular speed ( rad/s )</em>

<em>α = angular acceleration ( rad/s² )</em>

<em>t = elapsed time ( s )</em>

<em>θ = angular displacement ( rad )</em>

$\texttt{ }$

<u>Given:</u>

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

<u>Asked:</u>

average angular speed = ?

<u>Solution:</u>

<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>

$\theta = \frac{( \omega + \omega_o )}{2} t$

$\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0$

$\theta = 10 \times 4.0$

$\boxed {\theta = 40 \texttt{ rad}}$

$\texttt{ }$

<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>

$\texttt{average angular speed} = \theta \div t$

$\texttt{average angular speed} = 40 \div 4.0$

$\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Moment of Inertia : brainly.com/question/13796477
The Ratio of the Moments of Inertia : brainly.com/question/2176655

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

4 0

3 years ago

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