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2 years ago

Define the Earth's atmosphere

1 answer:

7 0

The blanket of gas on the surface of a planet or satellite. Note: The atmosphere of the Earth is roughly eighty percent nitrogen and twenty percent oxygen, with traces of other gases. (See ionosphere, stratosphere, and troposphere.)

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A bullet with a mass ????b=13.5mb=13.5 g is fired into a block of wood at velocity ????b=253vb=253 m/s. The block is attached to

Lorico [155]

Answer:

450 grams

Explanation:

Given:

mass of the bullet, m = 13.5 g = 0.0135 kg

velocity of the bullet, v = 253 m/s

spring constant of the spring, k = 205 N/m

Compression of the spring, x = 35.0 cm = 0.35 m

Now, the kinetic energy of the moving system (bullet + board) will tend to move the spring

thus,

let velocity of the system, V

now, applying the concept of conservation of momentum, we have

mv = (M + m)V

where,

M is the mass of the block

thus,

V = mv/(M + m)

now,

the kinetic energy of the system = (1/2)(M + m)V²

the kinetic energy of the system = (1/2)(M + m)(mv/(m + M))²

Energy gained by the spring = (1/2)kx²

now,

equating both the energies, we get

(1/2)(M + m)(mv/(m + M))² = (1/2)kx²

(mv)²/(m + M) = kx²

on substituting the values, we get

(0.0135 × 253)²/(0.0135 + M) = 205 × (0.35)²

11.66/(0.0135 + M) = 25.1125

M = 0.450 kg = 450 grams

8 0

3 years ago

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3

Advocard [28]

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

$a_c =\dfrac{v^2}{r}$

$2.32 =\dfrac{3.43^2}{r}$

$r =\dfrac{3.43^2}{2.32}$

r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

5 0

2 years ago

What is the change in velocity if the final velocity is 80 mph and the initial velocity is 20 mph? Is the object speeding up or

Dimas [21]

Answer: Δv=vf-vi=80mph-20mph=60mph

7 0

2 years ago

A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0

sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

$v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})$

Since the initial height is 0

Solving for $v_{o}$

$v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s$

5 0

3 years ago

The ___ model of the atom states that an electron's exact location within an atom can not be determined, but its probable locati

Dima020 [189]

The _quamtum mechanical_ model of the atom states that an electron's exact location within an atom can not be determined, but its probable location can be estimated within a three-dimensional region called an atomic orbital and that an electron's properties within an orbital can only be described by a set of mathematical values called a quantum number.

5 0

2 years ago