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hammer [34]
3 years ago
13

A) What minimum velocity must a roller coaster have such that the riders don’t fall out at the top of a loop with a radius of 12

.0 m? [3 marks]
B) What is the normal force a passenger of mass 60.0 kg experiences, If this same roller
coaster has a velocity of 13.0 m/s at the bottom of the loop? [5 marks]
Physics
1 answer:
kramer3 years ago
7 0

Answer: 100 miles per hour

Today, tubular steel tracks and polyurethane wheels allow coasters to travel over 100 miles per hour (160 km/h), while even taller, faster, and more complex roller coasters continue to be built. Hopefully i helped

Explanation:

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PLS HELP THIS IS WORTH A LOT OF POINTS ON MY TEST. WILL GIVE BRAINLIEST
Ede4ka [16]

Answer:

Explanation:

By conservation of energy, speed of the ball going up = speed of ball coming down with the ball stops at the top.

Because the gravity acceleration is constant, by symmetry, half of total time, 6/2 = 3s, is for going up and the last 3s for coming down.

Consider the last 3s when the ball drops from top to bottom, the initial velocity = 0 and acceleration = 10m/s^2

distance traveled = initial velocity * time + 1/2 * acceleration * time^2

= 0*3 + 1/2*10*3^2

= 5*9

= 45m

So maximum height of the ball is 45m.

7 0
3 years ago
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The saturnian moon io probably collided with
LUCKY_DIMON [66]
The Voyager and Pioneer flybys of the 1970s and 1980s provided rough sketches of Saturn’s moons. But during its many years in Saturn orbit, Cassini discovered previously unknown moons, solved mysteries about known ones, studied their interactions with the rings and revealed how sharply different the moons are from one another.
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3 years ago
In a softball game, a batter hits the ball at the velocity and angle shown
Kipish [7]
This problem is going to be pretty long to solve. So, prepare.

We’re interested in the change in our x position. So we have to break the velocity vector up into its components. Do cosine of 50 and then multiply by the magnitude of the velocity. I got 20.57m/s. That’s our initial velocity. And remember, horizontal acceleration is zero. The vertical acceleration, or any vertical component, has no effect on the horizontal components. In order to solve this problem, we want to utilize this equation:
Change in x-position = Vix*t

Let’s solve for time, which is dependent on the vertical components. The projectile will stop when it vertically hits the ground. Generally you want to use this equation for solving for time:
Yf = Yi + Viy*t + 1/2at^2
We didn’t solve for the vertical component yet, so let’s do that now. (Sine of 50)*(32) = 24.51m/s
Let’s now plug everything in:
0 = 0 + 24.5t - 4.9t^2
0 = 24.5t - 4.9t^2
0 = t(24.5 - 4.9t)
-24.5 = -4.9t
t = 5 seconds

The hard stuff is pretty much over. Put that 5 seconds into the other equation I said we wanted to use to solve the problem
Change in x-position (range) = (20.57)*(5)
= 102.85 meters

Answer B
5 0
4 years ago
What is described as the flow of charged particles?
rodikova [14]

Answer:

An electric current

Explanation:

8 0
3 years ago
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What changes when the force of gravity changes?
Semmy [17]

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

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