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Studentka2010 [4]
2 years ago
12

A 6.00-m long rope is under a tension of 600 N. Waves travel along this rope at 40.0 m/s. What is the mass of the rope?

Physics
1 answer:
MAVERICK [17]2 years ago
6 0

Answer:

2.25 kg

Explanation:

Length, L = 6 m

tension, T = 600 N

velocity, v = 40 m/s

The formula for the velocity is given by

v=\sqrt{\frac{T}{\mu }}

where, μ is mass per unit length

\mu =\frac{T}{v^{2}}

\mu =\frac{600}{40^{2}}

μ = 0.375

m / L = 0.375

m = 0.375 x 6 = 2.25 kg

Thus, the mass of rope is 2.25 Kg.

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Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate
Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J

Time required to burn the fat

t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s

Distance traveled by the bird

s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m

The bird will fly 455165.278 m

4 0
1 year ago
Read 2 more answers
Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌
NemiM [27]

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

(m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]

7 0
1 year ago
A car speeds up from 13 m/s to 23 m/s in 30 seconds. What is the acceleration of the car?
ra1l [238]
Well, <span>v = u + a×t is the equation.</span>
<span>
v: final velocity, which is 23 m/s in this equation.</span>
<span>u: initialo velocity = 13 m/s </span>
<span>a: acceleration = ? </span>
<span>t: time = 30s 
</span>
Your equation would be...

<span>23 = 13 + a×30 </span>
<span>a = (23 - 13) / 30 </span>
<span>a = 1 / 3 </span>
<span>a = 0.333 m/s</span>
7 0
2 years ago
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gayaneshka [121]

i think its d on edge

4 0
2 years ago
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A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at
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-- As she lands on the air mattress, her momentum is (m v)

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Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = <em>360 km-m/s </em>

The direction of the change is <em>up /\ </em>.

8 0
2 years ago
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