What type of boundary is shown below?

Ignoring air resistance, an object in free fall will fall d feet in t seconds, where d and t are related by the algebraic mode

Lostsunrise [7]

Explanation:

Given that,

An object in free fall will fall d feet in t seconds, where d and t are related by the algebraic model as :

$d=16t^2$ ..........(1)

(a) We need to find the time taken by the object to fall 1148 ft. Put this in equation (1) as :

$16t^2=1148$

$t=\sqrt{71.75}$

t = 8.47 seconds

(b) If the object is in free fall for 18.5 sec after it is dropped, then the height of the object is given by :

$d=16(18.5)^2$

d = 5476 ft

Hence, this is the required solution.

7 0

4 years ago

2. Which of the following is an example of acceleration.

N76 [4]

C
<span>Write at least 20 characters to explain it well.</span>

3 0

4 years ago

Read 2 more answers

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

•What power does a pump develop to lift 35 L of water per minute

guajiro [1.7K]

Answer:

$p = \frac{w}{t} \\ p = \frac{f \times d}{t} \\ p = \frac{m \times g \times d}{t} \\ p = \frac{35 \times 9.8 \times 110}{60.0} \\ p = 629w$

8 0

3 years ago

A 5 kg projectile is fired at an angle of 25o above the horizontal. Its initial velocity is 200 m/s and just before it hits the

lina2011 [118]

Answer:

The change in the mechanical energy of the projectile is 43,750 J

Explanation:

Given;

mass of the projectile, m = 5 kg

initial velocity of the projectile, u = 200 m/s

final velocity of the projectile, v = 150 m/s

The change in mechanical energy is calculated from the principle of conservation of energy;

ΔP.E = ΔK.E

The change in potential energy is zero (0)

0 = ΔK.E

ΔK.E = K.E₁ - K.E₂

ΔK.E = ¹/₂mu² - ¹/₂mv²

ΔK.E = ¹/₂m(u² - v²)

ΔK.E = ¹/₂ x 5(200² - 150²)

ΔK.E = 43,750 J

Therefore, the change in the mechanical energy of the projectile is 43,750 J

4 0

3 years ago