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jek_recluse [69]
3 years ago
6

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

uld the car have to accelerate so that a quarter placed on the back wall would remain in place?
Physics
1 answer:
Helga [31]3 years ago
6 0

Answer:3.23 m/s^2

Explanation:

Given

\mu_s =0.330

Frictional Force is balanced by force due to car acceleration

Frictional force F_s

F_s=ma_{min}

\mu_sN=ma_{min}

\mu_s\cdot mg=ma_{min}

a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2

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1 PHYSICS QUESTION - DEPENDENT AND INDEPENDENT VARIABLES. WILL GIVE BRAINLIEST!!!
nadya68 [22]

Answer:

The dependant variable is obvrioulsy going to be the temperature of the watch, and the independan variable is going to be the amount of water ebing poured into the cups.

Explanation:

The temperature is the dependant variable by how it is the thing being observed, or recorded. Your independant variables could be a few things from wha information I have but it could be either, the amount of water being poured into the cups, the temperature of the water being poured, or the amount of time between each new temperature of wather bing poured into the cups.

3 0
3 years ago
If a ball of mass 40 g moving with a speed of 200 m/s is brought to
Flura [38]

Answer:-400 n

Explanation:

8 0
2 years ago
The earth's orbital angular speed in rad/s due to its motion around the sun
Anna71 [15]

           (2π rad/year) x (1 yr/365 days) x (1 day / 86,400 seconds)

       =    (2π) / (365 x 86,400)        rad/sec

       =        0.000 000 2  radian/sec

       =          0.2 microrad/sec
5 0
3 years ago
An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H? An obj
Serhud [2]

Answer:

H = 171.90 m

Explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H = \frac{1}{2} gt^2    ......................1

here t is time and a is acceleration

so

we find t first

we know during time (t -1) s , it fall distance (H - 53.2) m

so equation of distance

H - 53.2 = \frac{1}{2} g (t-1)^2

H - 53.2 = \frac{1}{2} g (t^2-2t+1)

H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g     ................2

now subtract equation 2 from equation 1 so we get

H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)

53.2 = gt - \frac{1}{2} g

53.2 = 9.81 t - \frac{1}{2} 9.8

t = 5.92 s

so from equation 1

H = \frac{1}{2} (9.81)5.92^2

H = 171.90 m

5 0
3 years ago
Please help me with the following question:
pishuonlain [190]

Answer: a. 667N

b. 665N

c. 54.5N

Explanation:

a) on the surface of the earth

W = mg

W = 68 × 9.81

= 667N

b) at the top of Everest (8848 m above sea level).

W =mg × R²/(R + H)²

W = 667 × [6378²/(6378 + 8.848)²

W = 665N

c) has 2 1/2 times the radius of the earth

W = mg × R²/(R + H)²

W = 667 × R²/(R + 2.5R)²

W = 54.5N

3 0
2 years ago
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