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1 year ago

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Which statements about inertia and centripetal force are correct? Check all that apply. Inertia is always present. Inertia cause

s objects to continue moving in circles. Centripetal force is always present. Centripetal force pulls objects toward the center of a circle. Inertia and centripetal force cause circular motion.

2 answers:

likoan [24]1 year ago

7 0

Inertia is always present.

Centripetal force pulls objects toward the center of a circle.

Inertia and centripetal force cause circular motion.

allsm [11]1 year ago

6 0

Inertia IS always present. Inertia is NOT the force that causes objects to continue moving in circles, that is centripetal force. Centripetal force is NOT always present. Centripetal force DOES pull objects toward the center of a circle. <span> Inertia and centripetal force DOES cause circular motion. Thank you and eat sand fren ;)</span>

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An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?

Sergeeva-Olga [200]

By v^2 = u^2 + 2gh
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s

4 0

2 years ago

Read 2 more answers

Two light bulbs are operated at a potential difference of 110 V. Light bulb A produces 60 W of power and light bulb B produces 1

Over [174]

Answer:a

Explanation:

Given

Potential difference $V=110\ V$

Power of bulb A $P_A=60\ W$

Power of bulb B $P_B=100\ W$

If voltage is same for both the bulbs then Power is given by

$P=\dfrac{V^2}{R}$

$P_A=\dfrac{(110)^2}{R_A}$

$60=\dfrac{110^2}{R_A}$

$R_A=201.66\ \Omega$

similarly

$R_B=\dfrac{110^2}{100}$

$R_B=121\ \Omega$

$R_A>R_B$

so current in bulb A is smaller than B

Thus the 60 W bulb has a greater resistance.

Thus option (a) is correct

6 0

1 year ago

A boy stands at a certain distance from a large building and blows a whistle. After 2.3s he hears the echo of the sound. He move

ddd [48]

Answer:

I) 57.5 m

Il) 50 m/s

Explanation:

Given that a boy stands at a certain distance from a large building and blows a whistle.

After 2.3s he hears the echo of the sound. The speed V of the sound will be:

V = 2X / T

Where X is the distance covered

V = 2X / 2.3

V = 0.87X ...... (1)

He moves 50m towards the building and blows his whistle again; this time the echo reaches him after 2.0s.

V = (2×50) / 2

V = 50 m/s

Substitutes the V into equation 1

50 = 0.87X

X = 50 / 0.87

57.5 m

Therefore, the Boys original distance from the building is 57.5m and the

Speed of sound in air is 50m/s

6 0

2 years ago

The superhero Green Lantern steps from the top of a tall building. He falls freely from rest to the ground, falling half the tot

ddd [48]

Answer:

1) its initial velocity is zero, 2) the downward direction as positive

3) h = 25.66 m

Explanation:

This is a free fall exercise.

1) with falls, its initial velocity is zero and the acceleration is constant throughout the path and is equal to the acceleration due to gravity.

2) a widely used selection to estimate the downward direction as positive

3) We know that for the second part of the fall

y₀ -y = h/2 at t = 1 s

y = y₀ + v₁ t + ½ g t²

where v₁ is the initial velocity of this interval at the point y = h / 2

v₁ t = (y -y₀) - ½ g t²

v₁ = h / 2 - ½ g t²

v₁ = h/2 - g/2

now let's write the equation for the first interval

v₁² = v₀² + 2 g (y₁ - y₀)

in this interval v₀ = 0

v₁² = 2 g (y₁ -y₀)

v₁² = 2g h/2

we write our system of equations

v₁² = (h/2 - g/2)²

v₁² = (2g h / 2)

(h /2 - g/2)² = (2g h / 2)

h² / 4 - 2 g/2 h/2 + (g/2)² = g h

h² / 4 - g h/2 - g h + g²/4 = 0

h² - 3 g h + g² =0

h² - 29.4 h +96.04 = 0

we solve the quadratic equation

h = [29.4 ±√ (29.4² - 4 96.04)] / 2

h = [29.4 ± 21.91] / 2

h₁ = 25.66 m

h₂ = 3.75 m

As the system takes more than 1 S to fall, the correct answer for the height is h = 25.66 m

6 0

2 years ago

While attempting a landing on the moon, astronauts had to change their landing site and land at a spot that was 4 kilometers awa

lesya [120]

The closest answer would be C.
The choices given do not give the exact value.

To answer this, you just need to remember the main formula:

$d = Vit + \frac{1}{2}gt^{2}$

Where:
d = distance/displacement
g = acceleration due to gravity
t = time in flight
Vi = initial velocity.

With this formula, you derive all the formulas you need to look for certain components. You need to keep in mind of the following:

---if you are looking for a vertical component(y), you need to use values of vertical motion.

$dy = Viyt + \frac{1}{2}gt^{2}$

*Viy is always 0m/s at the beginning of a free-fall.

$dy = (0m/s)t + \frac{1}{2}gt^{2}$

$dy = \frac{1}{2}gt^{2}$

---If you are looking for a horizontal component(x), you need to use values of horizontal motion.

$dx = Vixt + \frac{1}{2}gt^{2}$

*g is always 0m/s² when taking horizontal motion into account.

$dx = Vixt + \frac{1}{2}(0m/s^{2})t^{2}$

$dx = Vixt$
--- time is the only value that is both vertical and horizontal.

Okay, let's get back to solving your problem. Let's see what your given is first:
dy = 137m (as long as it refers to height, it is vertical distance)
dx = 4km (the word, far or away usually indicates horizontal distance)
g = 1.63m/s²

The question is how fast was it going horizontally and we can derive it from our equation:

$dx = Vixt$

We use this because x means horizontal. But notice that we do not have time yet. So how are we going to solve this 2 variables missing? The key is that time is a horizontal and vertical component. Whatever time it took moving horizontally, it is the same vertically as well. So we use the vertical formula to derive time:

$dy = \frac{1}{2}gt^{2}$
$\frac{2dy}{g}=t^{2}$
$\sqrt{\frac{2dy}{g}} = \sqrt{t^{2}}$
$\sqrt{\frac{2dy}{g}} = t$

Now plug in what you know and solve for what you don't know:

$\sqrt{\frac{2(137m)}{1.63m/s^{2}}} = t$
$\sqrt{\frac{274m}{1.63m/s^{2}}} = t$
$\sqrt{168.098s^{2}}=t$
$12.965s = t$

The total time in flight is 12.965s.
Let's round it off to 13s.

Now that we know that, we can use this in the horizontal formula:
$dx = Vixt$
$4km = Vix(13s)$

Hold up! Look at the unit of the horizontal distance. It is in km but all our units are expressed in m so we need to convert that first.

1km = 1,000m
4km = 4,000m

Our new horizontal distance is 4,000m.

Okay, let's wrap this up by solving for what is asked for, using all the derived values.
$dx = Vixt$
$4,000m = Vix(13s)$
$\frac{4,000m}{13s} = Vix$
$308m/s= Vix$

The horizontal velocity is 308m/s.

Such a long explanation I know, but hopefully, you learned from it.

6 0

2 years ago