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Gre4nikov [31]
4 years ago
12

What is a benefit of fuel cell cars over standard cars?

Physics
1 answer:
Wittaler [7]4 years ago
5 0

Answer:

Fuel cells have a higher efficiency than diesel or gas engines.

Most fuel cells operate silently, compared to internal combustion engines.

Fuel cells can eliminate pollution caused by burning fossil fuels.

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A kid drops 1.2kg heavy ball vertically down. The ball hits ground and vertically bounce back. Speeds of the ball just before an
iVinArrow [24]

Answer:

8400 N

Explanation:

We are given that

Mass,m=1.2 kg

Initial speed,u=-8 m/s

Final speed,v=6 m/s

Time,\Delta t=2 ms=2\times 10^{-3} s

1 ms=10^{-3} s

We have to find the magnitude of the average force on the ball during this collision.

Change in momentum,=\Delta p=m(v-u)=1.2(6+8)=16.8 kgm/s

Average force,F_{avg}=\frac{\Delta p}{\Delta t}

Using the formula

F_{avg}=\frac{16.8}{2\times 10^{-3}}=8400 N

Hence, the magnitude of the average force on the ball during this collision=8400 N

8 0
4 years ago
If you divide a speed in miles per minute by 60, you get the same speed
Mariulka [41]
Yes so x multiply by the hour but u add 59 for each hour to get the exath speed per minute
6 0
3 years ago
Why do we need to conserve water?It is an unlimited resource.It is a limited resource.There is a lot of water.There is no need t
olchik [2.2K]

Answer:

conserving water is important because it keeps water pure and clean while protecting the inviorment

6 0
3 years ago
How many electrons can the first shell hold
Naily [24]
First shell hold up to 2 electrons.
8 0
4 years ago
What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is
QveST [7]
Missing detail in the text:
"<span>A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
</span>F=qE
<span>Considering the charge on the bead as a single point charge, the electric field generated by it is
</span>E=k_e  \frac{Q}{r^2}
with k_e = 8.99\cdot 10^9 Nm^2/C^2, Q=+25 nC=25 \cdot 10^{-9}C is the charge on the bead. We want to calculate the field at r=4.0 cm=0.04 m:
E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m
The proton has a charge of q=1.6\cdot 10^{-19}C, therefore the force exerted on it is
F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, m=1.67\cdot 10^{-27} kg, we have
F=ma
a= \frac{F}{m}= \frac{2.25\cdot 10^{-14} N}{1.67\cdot 10^{-27} kg}=1.35 \cdot 10^{13} m/s^2

The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
a=-1.35 \cdot 10^{13} m/s^2
6 0
3 years ago
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