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4 years ago

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At what point on a roller coaster is the kinetic energy increasing? As it is going down the hill. At the top of the hill. As it

is climbing the hill. At the bottom of the hill.

1 answer:

blagie [28]4 years ago

8 0

Answer:

As it is going down the hill

Explanation:

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Study the image of the moving car.

gayaneshka [121]

Answer:

While traveling downhill, the car’s potential is <u>increasing</u> and kinetic energy is <u>decreasing</u>

Explanation:

hope this helps!

6 0

2 years ago

Read 2 more answers

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.

wlad13 [49]

Answer:

12m/s

Explanation:

$v_f=v_o+at$

Let's call the velocity that the car maintains for 10 seconds $v_f_1$ , and the final velocity $v_f_2$ .

$v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s$

Hope this helps!

5 0

3 years ago

Minchanka [31]

Answer:

C

Explanation:

3 0

3 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

3 years ago

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is

mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate = 150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

$h= \frac{1}{2}at^2$

$h= \frac{1}{2}4.71\times4.96^2$

h= 58.012 m

h≅ 58 m

4 0

3 years ago