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zheka24 [161]
3 years ago
11

The public is not yet able to purchase cars powered by hydrogen fuel cells because engineers have to determin

Physics
1 answer:
Levart [38]3 years ago
3 0

Answer:

The options are

A.on racetracks

B.in real-world conditions

C.in flooded environments

D.on closed courses

The answer is B. In real world conditions

The public is not yet able to purchase cars powered by hydrogen fuel cells because engineers have to determin

how the cars perform based on real world conditions.

This will ensure they encounter the real and first hand experiences about the challenges and also the advantages associated with using this type of fuel.

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What is CHA-CHA-CHA.​
polet [3.4K]

Answer:

The cha-cha-cha, is a dance of Cuban origin. It is danced to the music of the same name introduced by Cuban composer and violinist Enrique Jorrin in the early 1950s. This rhythm was developed from the danzón-mambo

5 0
2 years ago
A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.4 m/s, a typical raindrop speed. How fast
kenny6666 [7]

The final velocity after the collision is 8.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, if we consider the system to be isolated (=no external unbalanced forces), the total momentum of the raindrop+mosquito must be conserved before and after the collision.

If the collision is perfectly inelastic, moreover, the raindrop and the mosquito stick together and travel at the same velocity v after the collision.

Mathematically:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 is the mass of the first mosquito

u_1 = 0 is the initial velocity of the mosquito

m_2 = 50 m_1 is the mass of the raindrop

u_2 = 8.4 m/s is the initial velocity of the raindrop

v is the final combined velocity of the raindrop+mosquito

Re-arranging the equation and substituting, we find:  

m_1 u_1 + 50 m_1 u_2 = (m_1 + 50 m_1) v\\50 m_1 u_2 = 51 m_1 v\\50 u_2 = 51 v\\v=\frac{50}{51}u_2 = \frac{50}{51}(8.4)=8.2 m/s

Learn more about momentum here:

brainly.com/question/7973509  

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4 0
3 years ago
HELP
blagie [28]

Answer:

A

Explanation:

hope its right

3 0
3 years ago
A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

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5 0
3 years ago
When was the first electric typewriter produced and by what company
LiRa [457]
In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
6 0
3 years ago
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