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Amanda [17]
3 years ago
11

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

Physics
1 answer:
SVEN [57.7K]3 years ago
6 0

Answer:

a) Knowns, initial speed v_{i}=13.0 m/s, final speed v_{f}=0 m/s and gravity due it is a constant g=9.8m/s^{2}

b) The maximum high reached by the dolphin is y_{max}=8.62 m

c) Total time is t=2.65s

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at 0m/s speed.

b) Second, now that we know final speed we use v_{f} =v_{i}-gt, as we clear for t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s.

Then we use y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2}  =8.62m

c)Third, finally we can use y=v_{i}t-\frac{1}{2} gt^{2}, as we know y=0m when the dolphin fall into the water again and v_{i} =13.0m/s, then we have 0=(13m/s)t-\frac{1}{2} (9.8m/s^{2}  )t^{2} is a quadratic form 0=t(13.0-4.9t) so we have t_{i}=0s and t_{f}=\frac{13}{4.90}  =2.65s

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3 years ago
Two factors that regulate (control) glandular secretion.
antiseptic1488 [7]

Answer:

The factors include age and puberty

Explanation:

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As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.

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3 years ago
(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is
Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H  = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = \dfrac{1}{2}mv^2

Using Conservation of energy

KE_i + PE_i + KE_f + PE_f

0 + m g H = \dfrac{1}{2}mv^2 + 0

v = \sqrt{2gH}

v = \sqrt{2\times 9.8 \times 105}

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0
2 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

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Answer:

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