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2 years ago

A lizard ran 3 meters from his rock to his friend's house. He ran

Physics

2 answers:

lbvjy [14]2 years ago

5 0

Answer:

4.5 meters

Explanation:

3 meters to his friend's house. +3

Halfway back. +1.5

3+1.5

=4.5

4.5 meters in total

Leya [2.2K]2 years ago

4 0

Answer:

Distance traveled, d = 4.5 meters

Explanation:

It is given that, a lizard ran 3 meters from his rock to his friend's house, distance covered by lizard, d₁ = 3 m

He ran back halfway and stopped, now the distance covered, $d_2=\dfrac{3}{2}=1.5\ m$

We need to find the distance traveled. The overall path covered by an object during its entire journey is called distance covered.

To find the distance covered by the lizard we need to simply add both distances as :

$d=d_1+d_2$

$d=3+1.5$

d = 4.5 meters

So, his distance traveled is 4.5 meters. Hence, this is the required solution.

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A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity

Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches = 0.0508 m

n = 1750 rpm

$H_{nom}=2hp=1491.4W$

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

$w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m$

$V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s$

$F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N$

$(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N$

$T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm$

$F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N$

$F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N$

$H_a=1491*1.25=1863.75W$

$n_f_s=\frac{H_a}{H_{nom}K_S }=1$

dip = $\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm$

7 0

2 years ago

a batter hits a homerun in which the ball travels 110m horizontally with no appreciable air resistance. If the ball left the bat

e-lub [12.9K]

If you know the formula for horizontal range, then finding the solution is immediate:

$r=\dfrac{{v_0}^2\sin2\theta}g$

$110\,\mathrm m=\dfrac{{v_0}^2\sin100^\circ}{9.8\,\frac{\mathrm m}{\mathrm s^2}}\implies v_0\approx33\,\dfrac{\mathrm m}{\mathrm s}$

7 0

2 years ago

A gymnast practices two dismounts from the high bar on the uneven parallel bars. during one dismount, she swings up off the bar

Fiesta28 [93]

Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.

g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.

First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s

Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s

The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.

Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward

Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward

The landing velocities differ by 0.43 m/s.

8 0

2 years ago

A scientist is examining an unknown solid. Which procedure would most likely help determine a chemical property of the substance

kifflom [539]

Answer:

b. exposing it to a flame to see if it catches on fire

Explanation:

The Procedure will most likely help to determine a chemical property of substance is : exposing material to a flame to see if it catches on fire Chemical property is the characteristic that a substance has that differentiate it from another substance. The most common charatcteristics that most scientists wanted to know are : - It's flammability - It's radioactivity - Its toxicity By throwing the object into fire, we will easily find out these 3 characteristics

Hence the correct answer is b. exposing it to a flame to see if it catches on fire.

3 0

2 years ago

Read 2 more answers

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc

USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

$m_A$ = mass of car A;