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3 years ago

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A student observes a shadow move across the sun during a solar eclipse. Which list represents the position of Earth, the sun, an

d the
moon during this event?

A.sun, Earth, moon

B.moon, sun, Earth

C.Earth, sun, moon

D.sun, moon, Earth

1 answer:

Liono4ka [1.6K]3 years ago

3 0

It goes sun moon earth the moon is blocking us from seeing the sun.

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If an object is an irregularly shaped solid and it is dropped into a graduated cylinder and it displaces 25 mL of water and has

zavuch27 [327]

Answer:

1250

Explanation:

7 0

3 years ago

Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

With the given values in the question the equation has one unknown v₀:

$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

Solving for t=1:

1) $v_0=y-y_0+\frac{g}{2}$

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) $E=\frac{1}{2}m{v_0}^2+mgy_0$

If h is the height of the tower, the energy on top of the tower:

3) $E=mgh$

Combining equation 2 and 3 and solving for h:

4) $h=\frac{{v_0}^2}{2g}+y_0$

Combining equation 1 and 4:

$h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0$

4 0

3 years ago

A racecar is traveling at a speed of 80.0 m/s on a circular

Simora [160]

A :-) F = mv^2 by r
Given - m = 500 kg
v = 80.0 m/s
r = 450 m
Solution -
F = mv^2 by r
F = 500 x (80)^2 by 450
F = 500 x 6400 by 450
( cut 500 and 450 because 5 x 100 = 5 , 5 x 90 = 450 and also 90 and 6400 because 90 x 70 = 6400 )
F = 100 x 70
F = 7000

.:. The centripetal force is 7000 N

7 0

3 years ago

Read 2 more answers

A remote-controlled boat traveled 18 feet heading west and 24 feet heading south, totaling 42 feet over the course of 10 seconds

Dominik [7]

Answer:

3 ft/s

Explanation:

given,

distance in west, d₁ = 18 ft

distance in south, d₂ = 24 ft

time taken to travel 42 ft = 10 s

speed of boat in north = x ft/s

Speed of the boat when heading west = (x - 3) ft/s

$Average\ speed = \dfrac{total\ distance}{total\ time}$

$v_{avg}= \dfrac{42}{10}$

$v_{avg}= 4.2\ ft/s$

again using the above formula

now, total time

t = t₁ + t₂

$10 = \dfrac{18}{x-3} + \dfrac{24}{x}$

on simplifying

10 x² -72 x + 72 = 0

On solving the above equation we get

x = 6 ft/s

Hence, speed of boat in west direction, 6 -3 = 3 ft/s

speed of boat in south direction = 6 ft/s

3 0

3 years ago

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$

Solving for initial velocity (when the boulder left the volcano):

$v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}$

$v_i=101.44 \frac{m}{s^{2}}$

3 0

3 years ago