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3 years ago

Heavenly body.

Physics

1 answer:

kicyunya [14]3 years ago

7 0

Answer:

768678677887667

Explanation:

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What is the safest way to enter a body of water? Enter with you feet first. Dive in.

zimovet [89]

Answer: feet first

Explanation:

It is always the safest to enter with your feet first. You can become paralyzed or injure you neck if you dive into a body of water that is too shallow. Therefore its always best to dive feet first instead.

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3 years ago

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A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

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3 years ago

astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o

devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ $T=2 \pi\sqrt{\frac{m}{k} }$

or,

⇒ $m=\frac{k T^2}{4 \pi^2}$

On substituting the values, we get

⇒ $=\frac{280\times (3)^2}{4 \pi^2}$

⇒ $=\frac{280\times 9}{4\times (3.14)^2}$

⇒ $=68.83 \ kg$

(b)

The speed of the string will be:

⇒ $\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2$

then,

⇒ $v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }$

On substituting the values, we get

⇒ $=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }$

⇒ $=\sqrt{\frac{280(0.16-0.04)}{63.83} }$

⇒ $=\sqrt{\frac{280\times 0.12}{63.83} }$

⇒ $=0.725 \ m/s$

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Tcecarenko [31]

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