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jonny [76]
3 years ago
5

How much force would a 55kg person experience on a roller coaster if they travel through a loop with a radius of 55kg at a speed

of 14.1 m/s?

Physics
1 answer:
Inessa [10]3 years ago
5 0

Answer:

200 N

Explanation:

For a body moving in uniform circular motion, the force acting on it will be <em>centripetal force</em> and its direction is <em>radially inward</em> , pointing to the center.

The radially inward acceleration, or the centripetal acceleration is given by :

                                          a = v² / r

           where v is the speed at which the body is moving and r is the radius of the circle

Given-

m = 55kg

v = 14.1 m/s

r= 55m

We know that F = ma

⇒   F = m (  v²/ r )

⇒ F = 55 x 14.1 x 14.1 / 55

⇒ F =14.1 x 14.1 = 200 N

∴ <em>The force acting is 200 N</em>.

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Abigail runs one complete lap (400m) around the track, while Gabi runs a 50 meter dash in a straight line. Which runner had a gr
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Answer:

\boxed {\boxed {\sf Gabi \ had \ greater \ displacement}}

Explanation:

Displacement is the change in the position of an object.

Abigail runs a complete lap around the track, which is 400 meters. Even though she ran 400 meters, she has <u>no displacement.</u> If she starts and ends at the same spot, there is no change in position.

Gabi runs a 50 meter dash in a straight line. Gabi has <u>50 meters of displacement</u>. She runs in a straight line and is 50 meters away from where she began.

While Abigail ran the farther distance, <u>Gabi had the greater displacement.</u>

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2 years ago
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A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go ?
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So, this is a problem where the accleration is not provided, since it is implied.  The only acceleration is acceleration due to gravity (9.8 m/s)


The equation we will use for this problem is V^2 =V_{0}^2 + 2a (X-X_0)

V is the final velocity, V₀ is the initial velocity, a is the acceleration, X is the final height, and X₀ is the starting height.


We can assume that the ball starts on the ground since no height is given, so now we plug our numbers in.

We will use 0 as the final velocity, since the ball will stop moving upwards when it is the highest.  We will use -9.8 since that is the acceleration due to gravity and we will use 22m/s as V₀ since that is the starting velocity.

V^2 =V_{0}^2 + 2a (X-X_0)\\0^2 = 22^2 + 2\times-9.8(X-0)\\0=484-19.6x\\-484=-19.6x\\24.69387755 = x\\x\approx24.69


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4 0
3 years ago
A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with constant speed of 3.50m/s. The coefficient of kin
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Answer:

3.125 m

Explanation:

We are given that

Mass of box=m=11.2 kg

Speed of box=u=3.5m/s

Coefficient of kinetic friction=\mu_k=0.2

Final velocity,v=0

a.We have to find the horizontal force applied by worker to maintain the motion.

According to question

Horizontal force=F=f=\mu_kmg

g=9.8m/s^2

Substitute the values

Horizontal force=F=0.2\times 11.2\times 9.8=21.95 N

b.According to work-energy theorem

W=\frac{1}{2}mv^2-\frac{1}{2}mu^2

-\mu mg s=\frac{1}{2}(11.2)(0)^2-\frac{1}[2}(11.20)(3.5)^2

-\mu mg s=-\frac{1}{2}(11.2)(3.5)^2

0.2\times (11.2)\times 9.8\times s=\frac{1}{2}(11.2)(3.5)^2

s=\frac{11.2\times (3.5)^2}{2\times 0.2\times 11.2\times 9.8}

s=3.125 m

Hence, the box slide before coming to rest=3.125 m

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