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natta225 [31]
3 years ago
10

A basketball rolls without slipping (starting from rest) down a ramp. If the ramp is sloped by an angle of 4 degrees above the h

orizontal and the ball rolls for a distance of 60 m along the ramp, what is its speed after rolling this far?The moment of inertia of a basketball is 1.
Physics
1 answer:
slavikrds [6]3 years ago
4 0

Answer:

11.7 m/s

Explanation:

To find its speed, we first find the acceleration of the center of mass of a rolling object is given by

a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²

a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)

  = 9.8 m/s²sin4(1 + 2/3)

  = 9.8 m/s²sin4 × (5/3)

  = 1.14 m/s²

To find its speed v after rolling for 60 m, we use

v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m

v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s

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008 (part 3 of 4) 3.0 points
motikmotik

Car A take a time of 2.55hr and car B take a time of 2.14 hr

We know that distance divide by time is speed

here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr​

so speed=distance/time

s=d/t

t=d/s

=189/74

=2.55hr

In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr  

s=d/t

t=d/s

=199.8/93

=2.14hr  

so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h

learn more about Speed here brainly.com/question/13943409

#SPJ9

5 0
1 year ago
One property that makes electromagnetic waves differ from other types of waves is that they can
Flauer [41]
travel through a vacuum at the speed of light. Other waves need a medium; sound waves need molecules that vibrate.
5 0
3 years ago
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi
True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0
3 years ago
1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic​
Korvikt [17]

51.448 g is the required answer!

8 0
3 years ago
A boat heads due North for 3.00 km to reach a red buoy. It then changes direction and heads in a direction of 30.0 degrees South
Helga [31]

Position of B :  

x = 4.66*cos 30 = 4.036  

y = 3-4.66*sin 30 = 3-2.33 = 0.67  

BC = √y^2+(x-1)^2 = √0.67^2+3.036^2 = 3.109  

heading = arctan y/(x-1) = arctan 0.67/(3.036) = 12.44° south of west

hope this helps :)

6 0
3 years ago
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