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maksim [4K]
4 years ago
6

Two identical cars, one on the moon and one on earth, are rounding banked curves at the same speed with the same radius and the

same angle. the acceleration due to gravity on the moon is 1/6 that of earth. how do the centripetal accelerations of each car compare?
a.the centripetal acceleration of the car on earth is less than that on the moon.

b.the centripetal acceleration of the car on earth is greater than that on the moon.

c.the centripetal accelerations are the same for both cars.

d.this cannot be determined without knowing the radius and the
Physics
1 answer:
Allisa [31]4 years ago
7 0
D because you are not given enough info

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As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh
Leya [2.2K]

<u>Solution and Explanation:</u>

The following calculation is made in order to find out the area and the final velocity vector.

Using the given information and the data in the question,  

m1u1 + m2u2 = (m1 + m2)  multiply with v

40000 multiply with ( -20i ) + 60000 multiply with ( 10j ) = 100000 multiply with v

  Therefore, v = -8i + 6j

That is  |v| = 10 knots towards the 36.86 degree north of the west

6 0
3 years ago
A rope connects boat A to boat B. Boat A starts from rest and accelerates to a speed of 9.5 m/s in a time t = 47 s. The mass of
Contact [7]

Answer: 339.148N

Explanation:

Data

Time (t) = 47s

U = 0m/s

V = 9.5m/s

Mass of B = 540kg

Frictional force on B = 230N

Both boats are connected so if A moves, B moves too.

Acceleration of boat A =?

Using equation of motion,

V = u + at

9.5 = 0 + a*47

a = 9.5 / 47

a = 0.2021 m/s²

The force required to accelerate boat B since it's the same force moving both boats =?

F = Mass * acceleration

F = 540 * 0.2021 = 109.14N

A frictional force of 230N exists on boat B

Total force (Tension) = frictional force + normal force = (109.15 + 230)N = 339.148N

7 0
3 years ago
20 points, just need a basic understanding. You are on a mystery planet, what you know is that from a height of 10.0 meters, a d
ICE Princess25 [194]

(since you asked for basic understanding only, I am not including actual calculations. Please let me know in the comments section if you wish to verify your solution(s))

For (b): Use the formula for distance (s) made during an accelerated motion:

s = \frac{1}{2}at^2+ v_0t+s_0= \frac{1}{2}at^2= \frac{1}{2}gt^2

with v_0 and s_0 being the initial velocity and distance, both 0 in this case, and with "a" denoting the acceleration, in this case  solely due to gravitational acceleration so: "g."

You are given the distance made, namely 10 m, and the duration t (0.88s) and so using the formula above you can solve for g.

For (c), to determine the final velocity at time 0.88s use the formula for the instantaneous velocity of an accelerated motion

(velocity at time t) = (acceleration) x (time)  

again, with acceleration due to gravity, i.e., a = g and with g as determined under (b).  

If my calculation is correct, this mystery planet could be the Jupiter.


7 0
4 years ago
What is true of a reaction that has reached equilibrium?
grin007 [14]
When a reaction has reached equilibrium, the rate of the forward reaction<span> equals the rate of the reverse </span>reaction<span>, such that the concentrations of reactants and products remain fairly stable, in a chemical </span>reaction<span>. Hope this answers the question. Have a nice day.</span>
4 0
4 years ago
_____ variables are manipulated by the experiments
nikitadnepr [17]

Answer:

dependent variables

Explanation:

dependent varibeles are the thing you're measuring and independent variables are the thing you change in the exeriment to get a different dependent variable.

may I get brainliest please? :)

3 0
4 years ago
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