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Ne4ueva [31]
3 years ago
12

A 52.0 kg skateboarder does a trick on a half-pipe. During the trick, he reaches a speed of 8.90 m/s. The radius of the half-pip

e ramp is 3.00 meters. What is the minimum centripetal force necessary to keep the skater on the ramp?
Physics
1 answer:
Lyrx [107]3 years ago
3 0
A= (8.9)^2/3=26.4
f=26.4*52=1372.97N
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Answer:

True

Explanation:

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3 years ago
. Calcula el módulo del vector resultante de dos vectores fuerza de 9 [N] y 12 [N] concurrentes en un punto o, cuyas direcciones
ycow [4]

Answer:

a) 20.29N

b) 19.42N

c) 15N

Explanation:

To find the magnitude of the resultant vector you can consider an axis in the middle of the vector, from which you can calculate the components of the vectors by using the angles given:

a) for 30°

F_1=[9cos(15\°)\hat{i}+9sin(15\°)\hat{j}]N\\\\F_1=[8.69\hat{i}+2.32\hat{j}]N\\\\F_2=[12cos(15\°)\hat{i}-12sin(15\°)\hat{j}]N\\\\F_2=[11.59\hat{i}-3.10\hat{j}]N\\\\F=F_1+F_2=20.28N\hat{i}-0.78N\hat{j}\\\\|F|=\sqrt{(20.28N)^2+(0.78N)^2}=20.29N

F = 20.29N

b)  for 45°

F_1=[9cos(22.5\°)\hat{i}+9sin(22.5\°)\hat{j}]N\\\\F_1=[8.31\hat{i}+3.44\hat{j}]N\\\\F_2=[12cos(22.5\°)\hat{i}-12sin(22.5\°)\hat{j}]N\\\\F_2=[11.08\hat{i}-4.59\hat{j}]N\\\\F=F_1+F_2=19.39N\hat{i}-1.15\hat{j}\\\\|F|=\sqrt{(19.39N)^2+(1.15N)^2}=19.42N

F= 19.42N

c) for 90°

for this case you can consider that the direction of both vectors are the y and x axis of the Cartesian plane:

F_1=9N\hat{i}\\\\F_2=12N\hat{j}\\\\F=\sqrt{(9N)^2+(12N)^2}=15N

F=15N

6 0
3 years ago
Can someone please help me I don’t understand this
jolli1 [7]
I think it’s a because it the only reasonable answer
4 0
3 years ago
Gary saved dimes and nickels at a ratio of 5:7. Then, he saved 20% more dimes and 3 times the original number of nickels. If he
TiliK225 [7]

Answer:

i think 53 or 56

Explanation:

cause i cross multiplied and got that

8 0
3 years ago
Read 2 more answers
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