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2 years ago

A bullet is fired horizontally at 343 m/s from the top of a building where height is 37.3 m. The built will be

Physics

1 answer:

Nikitich [7]2 years ago

5 0

Answer:

Explanation:

A classic projectile motion question. Horizontal motion does not matter whatsoever in the context of this equation as we are looking at the vertical component of velocity only. Since the question states that it is fired horizontally, there is no vertical component of velocity. Thus, v0y = 0.

Assuming the projectile is on the Earth, we can use the free-fall acceleration, 9.8 m/s^2.

The question gives us Δy, which is 37.3m.

Using the second kinematics equation, Δy = v0t + (at^2)/2, we can substitute in the values.

37.3 = 0 + 0.5(9.8)t^2

37.3 = 4.9t^2

7.61224489796 = t^2

t = 2.75902970226 seconds.

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How know that rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the arkose sandstone

svp [43]

The rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the Arkose sandstone because the Amadeus basin were made up of marine and non-marine sedimentary rocks which are softer compared to quarts which make up mostly the Arkose sandstone.

7 0

3 years ago

Read 2 more answers

A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a

saw5 [17]

Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

Boyle's Law, (P₁V₁ =P₂V₂)
Charles' Law (V₁/T₁ =V₂/T₂)
Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

where P = Pressure, T = Absolute temperature, V = Volume occupied

The volume of the system remains constant,

So, P₁/T₁ =P₂/T₂

a) $\frac{5}{278} =\frac{P_2}{313} \\\\P_2=\frac{5*313}{278}\\ P_2 = 5.63 atm$

7 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs.

Brut [27]

Answer:

5 percent = normal matter

68 percent = dark energy

27 percent = dark matter

Explanation:

3 0

2 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

4 0

3 years ago

What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall

andrey2020 [161]

Answer:

$D_{s}$ ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere $Is_{cm}$ = 2/5 M R²

Spherical shell $Ic_{cm}$ = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

I = $I_{cm}$ + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

D = 2R

Ic = $Ic_{cm}$ + M D²

Ic = 2/3 MR² + M (2R)²

Ic = M R² (2/3 + 4)

Ic = 14/3 M R²

The sphere

Is = $Is_{cm}$ + M [ $D_{s}$ ²

Is = Ic

2/5 MR² + M $D_{s}$ ² = 14/3 MR²

$D_{s}$ ² = R² (14/3 - 2/5)

$D_{s}$ = √ (R² (64/15)

$D_{s}$ = 2,066 R

3 0

3 years ago