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VladimirAG [237]
3 years ago
13

It’s upside down, I know.

Physics
1 answer:
Paladinen [302]3 years ago
3 0
Where's the question?
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Nancy rides her bike with a constant
Sloan [31]

Answer:

2 and a half hours

Explanation:

Nancy rides her bike at 10 miles per hour. After the first hour, she would have ridden 10 miles. After the second hour, she would hvae ridden 20 miles. She needs to ride 5 more miles, which is half of her 10 miles that she rides every hour. Since it is half of her speed of miles per hour, it will take her another half an hour to ride the 5 miles. Added together, Nancy takes 2 and a half hours.

7 0
3 years ago
What is the smallest possible number of products in a decomposition reaction?
Vaselesa [24]
The smallest number of products in a decomposition reaction is two because there is 1 reactant that breaks down and when it does break down there must be two products. It is generally more, though.
3 0
3 years ago
A 15 cm3 block of gold weighs 2.8 N. It is carefully submerged in a tank of mercury. One cm3 of mercury weighs 0.13 N.
BaLLatris [955]
The density of gold is  2.8 N / 15 = 0.18 units
The density of mercury is 0.13/1 = 0.13 units

Since the density of gold is more than density of mercury, it will sink.
Since the gold will sink, it will displace mercury, but less than its own volume
5 0
3 years ago
A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity
Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

\omega_i = 2\pi f_1

\omega_1 = 2\pi(\frac{610}{60})

\omega_1 = 63.88 rad/s

similarly final angular speed is given as

\omega_f = 2\pi f_2

\omega_2 = 2\pi(\frac{837}{60})

\omega_2 = 87.65 rad/s

angular acceleration of the turbine is given as

\alpha = 5.9 rad/s^2

now we have to find the angular displacement is given as

\theta = \omega t + \frac{1}{2}\alpha t^2

\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)

\theta = 234.62 radian

3 0
3 years ago
The Zero Gravity Research Facility at the NASA Glenn Research Center includes a
denis23 [38]

Answer: (a) t = 5.44 sec

(b) vf = 53.31 m/s

(c) s = 5.0m

Explanation: from the question, given data

the Height of the tower, h = 145m

from question

(a)

the initial velocity, v₁ = 0 m/s

s = v₁t + 1/2 gt²

-145 m = 0(t) + 1/2 (-9.8t²)

  t² = 145/4.9

   t² = 29.59

    t = 5.44 sec

(b)

the speed of the sphere at the bottom of the tower is

vf² = vi² +2as

vf² = 0 + 2(-9.8 × -145)

vf² = 2842

vf = 53.31 m/s

(c)

when caught, the sphere experiences a deceleration of;

   a = -29.0g

the time it would take to decelerate becomes;

vf = vi + at

0 = (53.31) + (-29 ×9.8)t

where t = 53.31 / 284.2

t = 0.1876 sec

∴ the distance travelled during the deceleration becomes;

vf² = vi² + 2as

s = (vf² - vi²) / 2a

s = (0 - 53.31²) / 2×-29×9.8

s = -2841.9561 / -568.4

s = 4.99 ≈ 5.0m

i hope this helps, cheers

4 0
3 years ago
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