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xxMikexx [17]
3 years ago
11

Consider a river flowing toward a lake at an average velocity of 3 m/s. the river height is 90m above the lake. what is the tota

l mechanical energy of the river water per unit mass in kj/kg?
Physics
1 answer:
strojnjashka [21]3 years ago
6 0

Kinetic energy per unit of mass is

K=\frac{v^{2} }{2}

Given, v=3m/s^{2}

Therefore,

K=\frac{(3^{2} m/s^{2} )^2}{2}

K=4.5 J/kg

Now potential energy per unit mass is

p=g\times h

Given, h=90 m

Therefore,

p= 9.8m/s^2 \times 90

p=882.9 J/kg

Thus, total mechanical energy of the river water per unit mass is

T=K+p=(4.5+882.9)J/kg

T=887.9 J/kg

OR

T=0.887 kJ/kg

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How does inertia affect a person who is not wearing a seatbelt during a collision?
Vadim26 [7]
When someone fails to wear a seat belt the passenger becoming a projectile the force a person will be subjected to for a passenger weighting 100 pounds, and the car is traveling at 60 mph would be the same as 6000 pounds. thats like hitting a brick wall. so is other words put ur seat belt on. 

hope this helped (:
4 0
3 years ago
How many millimeters are 120 meters ? Help ! Please <3
blsea [12.9K]

Answer:

120,000

Explanation:

Millimeters to meters calculation-

Multiply by 1,000.

120 x 1,000 = 120,000.

This is the correct answer and formula.

Hope this helps!

8 0
3 years ago
Read 2 more answers
Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len
Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

4 0
2 years ago
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at
Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

<span>
We can solve this problem by using the formula:</span>

v^2 = v0^2 + 2 a d

 

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

<span>a = 662.625 km/h^2</span>

6 0
3 years ago
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