Wilbur ran 1-kilometer. Then he ran 500 meters. How many meters did Wilbur run all together?

An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminu

Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

$s=-0.9t^2+36t$

Differentiating with respect to time we get

$\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36$

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

$0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s$

b) The object will reach the highest point after 20 seconds

$s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m$

c) Highest point the object will reach is 360 m

$s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0$

$\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s$

d) Time taken to strike the ground would be 20+20 = 40 seconds

$[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s$

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of $s=ut+\frac{1}{2}at^2$

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

4 0

3 years ago

A boat travels north across a river at a velocity of 22 meters/second with respect to the water. The river's velocity is 2.2 met

MArishka [77]

The magnitude of the resultant is

√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .

The direction of the resultant is

tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .

7 0

3 years ago

An adult human is 60 percent water; a third of this water is in extracellular fluid, and 20 percent of extracellular fluid is in

Yanka [14]

Answer:

32 pounds

Explanation:

The amount of water in the 200 pound person is

200 * 60% = 200*0.6 = 120 pounds

Of the 120 pounds, a third of this is extracellular fluid, the amount of extracellular fluid is

120 / 3 = 40 pounds

20 % of this is in the blood, which amounts to

40 * 20% = 40 * 0.2 = 8 pounds

The rest is interstitial fluid, which is

40 - 8 = 32 pounds

5 0

3 years ago

An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit

nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

5 0

3 years ago

Which formula can be used to solve problems related to the first law of thermodynamics?

docker41 [41]

Answer: The formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

$\Delta U=Q+W$

where, Q = heat energy

W = work energy

$\Delta U$ = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For $\Delta U$ : When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

8 0

3 years ago

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