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3 years ago

6

Name a common, everyday circumstance in which objects are under constant acceleration and describe why those objects are under c

onstant acceleration (2 points)

2 answers:

asambeis [7]3 years ago

8 0

Answer:

Everything on the Earth is under the influence of gravity, which is a constant acceleration towards the center of the Earth with the magnitude of 9.81 m/s^2. When you drop something, it falls to the ground. If you bungee jump, you fall with a constant acceleration of -9.81 m/s^2 until the ropes pull you.

MatroZZZ [7]3 years ago

6 0

Answer:

Your age, because you never stop aging even when you're dead you still aging even though you're not alive.

Hope this helped

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On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial

swat32

Explanation:

We know that that the range of the ball on the earth

$R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}$

therefore, range of the ball on moon

$R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}$

$R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}$

therefore,

$R_{moon}=6R_{earth}$

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0

3 years ago

Mga tanong 1. Ano ang nangyari sa lobo? 2. Sa iyong palagay, saan napunta ang pumutok na lobo? 3. Kapag pumutok ang lobo ano ang

gavmur [86]

Answer:

1. Pumutok ang Lobo

2.basurahan

3.may kwentong po ba to paano ko masagot ng wala naman akong mababasang kwento :) :) :)

4 0

3 years ago

An example of a double reaction is

butalik [34]

C. is the only double reaction here given that a double replacement reaction involves two compounds that exchange previous components, and C is the only solution with two compounds present

8 0

4 years ago

The Problems: 1. Xavier starts at a position of 0 m and moves with an average speed of 0.50 m/s for 3.0 seconds. He normally mov

NemiM [27]

Answer:

(1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

Explanation:

(1). Given that,

Initial position = 0 m

Average speed = 0.50 m/s

Time = 3.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Where, $x_{f}$ = final position

$x_{i}$ = Initial position

t = total time

Put the value into the formula

$0.50=\dfrac{x_{f}+0}{3.0}$

$x_{f}=0.50\times3.0$

$x_{f}=1.5\ m$

(2). Given that,

Initial position = 0 m

Average speed = 0.75 m/s

Time = 4.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$0.75=\dfrac{x_{f}+0}{4.0}$

$x_{f}=0.75\times4.0$

$x_{f}=3\ m$

(3). Given that,

Average speed = 1.25 m/s

Time = 3.0 sec

Initial position = 1.0 m

We need to calculate the final position

Using formula of average speed

$v=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$1.25=\dfrac{x_{f}+1.0}{3.0}$

$x_{f}=1.25\times3.0-1.0$

$x_{f}=2.75\ m$

(4). Given that,

Average speed = 1.25 m/s

Distance = 100 m

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100}{1.25}$

$t=80 sec$

(5). Given that,

Average speed = 5 miles/hr

Suppose, distance = 25 miles

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{25}{5}$

$t=5\ hr$

Hence, (1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

5 0

4 years ago

A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so

joja [24]

ANSWER:100dB

f<em>rom the sound intensity level,the sound intensity is calculated as:</em>

<em /> $\beta$ <em>₁=</em> $\beta$ <em>=(10dB)㏒₁₀(l₁/l₀)</em>

<em>inserting numbers:</em>

<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>

<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>

<em>can be used to solve for l₁ and get</em>

<em> l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>

<em>since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>

<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>

<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>

<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>

<em> </em> $\beta$ <em>₂=</em> $\beta$ <em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em> </em> $\beta$ <em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em /> $\beta$ <em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

<em />

3 0

4 years ago