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Nadusha1986 [10]
2 years ago
15

Linda, a biker, is moving along a circular path at a constant speed of 10 km/h. In a neighboring arena Kevin, another biker, is

moving along a circular path at a constant speed of 12 km/h. If the acceleration of the two bikers is the same, what can you infer?
A) The centripetal force on Kevin is higher than that on Linda.
B) The centripetal force on Linda is higher than that on Kevin.
C) Kevin’s circular path has a bigger radius than Linda’s path.
D) Linda’s circular path has a bigger radius than Kevin’s path
Physics
1 answer:
Ulleksa [173]2 years ago
8 0

Answer:

the answer is  C

Explanation:

i took the test just now so i am glad i can help someone in the future❣❣❣❣

<u><em>Also here is the explanation</em></u>

Note that the magnitude of acceleration (a) is given by a =  

v2

R

, where v is the speed and R is the radius. Because acceleration is the same for both bikers, the ratio 

v2

R

, also has to be the same for both. Given that Kevin’s speed is higher, the radius of his path must also be higher to maintain the same ratio. Therefore, Kevin’s circular path has a bigger radius than Linda’s path.

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An astronaut goes to Mars to do some experiments. Explain why her mass stays the same but her weight changes.
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3 years ago
A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
3 years ago
How do you solve 0.004 dm + 0.12508 dm?
Effectus [21]
0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.  

The thing could be a glass of water, a sheet of paper,
a pound of ground beef, a gallon of gas, or a snowball.  
In this problem, it just happens to be a dm. 
7 0
3 years ago
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