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Tanzania [10]
3 years ago
11

What is the average velocity of an arrow that travels 12 m [E] in 0.15 s?

Physics
1 answer:
sergij07 [2.7K]3 years ago
6 0
Velocity is defined as Distance divided by Time.
In other words, V = D/T.

Now that we have our formula, we can solve.
Let's plug in the numbers we have.

We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.

Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.

Your velocity is 80 m/s [E]

I hope this helps!
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2 years ago
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A spherical shell of radius 3.59 cm and a cylinder of radius 7.22 cm are rolling without slipping along the same floor. The two
vampirchik [111]

Answer:

(ω₁ / ω₂) = 1.9079

Explanation:

Given

R₁ = 3.59 cm

R₂ = 7.22 cm

m₁ = m₂ = m

K₁ = K₂

We know that

K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²

if

v₁ = ω₁*R₁

and

I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²

∴    K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁²   <em>(I)</em>

then

K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²

if

v₂ = ω₂*R₂

and

I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²

∴    K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²   <em>(II)</em>

<em>∵   </em>K₁ = K₂    

⇒   0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒  ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

⇒  (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²

⇒  (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)

⇒  (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²

⇒  (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒  (ω₁ / ω₂) = 1.9079

8 0
3 years ago
During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
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Answer:

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(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

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A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

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3 years ago
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Answer:2.5

Explanation:

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lapo4ka [179]

Answer:  100 m/s^2

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Explanation:

50N = 50 kg*m/s^2

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F=ma

a = F/m

a = (50 kg*m/s^2)/(0.5 kg)

a = 100 m/s^2

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