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Artemon [7]
3 years ago
12

A bee flies forward at 4.9m/s for 48s , lands on a flower and stays there for 28s , then flies back along its previous route at

5.1m/s for 38s . What is the total displacement of the bee? Round your answer to the nearest thousandth, if necessary.
Physics
1 answer:
Rom4ik [11]3 years ago
3 0

"Displacement" is the distance and direction between the start-point and the end-point, regardless of the route taken on the way.

From this definition, it's easy to see that the bee's displacement at the end of the adventure is zero.

The bee's distance and average speed could also be calculated using the given information, but are not requested.

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A 215-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a
Misha Larkins [42]

Answer:

303.9481875 N

Explanation:

t = Time taken = 2 seconds

F = Force

r = Radius = 1.5 m

I = Moment of Inertia

\alpha = Angular Acceleration

Torque

\tau=F\times r

\tau=I\times \alpha

\\\Rightarrow F\times r=I\times \alpha\\\Rightarrow F=\frac{I\times \alpha}{r}

Angular velocity

\omega=rev/s\times 2\pi\\\Rightarrow \omega=0.6\times 2\pi\\\Rightarrow \omega=3.76991\ rad/s

Angular acceleration

\alpha=\frac{\omega}{t}\\\Rightarrow \alpha=\frac{3.76991}{2}\\\Rightarrow \alpha=1.88495\ rad/s^2

I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}215\times 1.5^2\\\Rightarrow I=241.875\ kgm^2

F=\frac{I\times \alpha}{r}\\\Rightarrow F=\frac{241.875\times 1.88495}{1.5}\\\Rightarrow F=303.9481875\ N

The magnitude of the force to stop the merry-go-round is 303.9481875 N

3 0
3 years ago
ASAP need physics help please:)
boyakko [2]

Answer:

plantation of force of the earth acting on 15 kg of object on free fall the acceleration of free fall said about it now it is year 15 kg then it becomes 1.53 Newton

3 0
2 years ago
Order the seven basic spectral types from hottest to coldest.
vfiekz [6]

Answer:

There are basically seven spectral types which are used for classifying the stars and it is arranged in the order of increasing temperature from the hottest temperature to the coldest division are: O, B, A, F, G, K, and M. These categories are basically divided into the sub category in the spectral where O is the hottest type and M is the coldest type.  

8 0
2 years ago
Bob is studying waves in science and learns that the energy of a wave is directly proportional to the square of the waves amplit
Mrrafil [7]
For a simple harmonic motion energy is given with:
E=\frac{1}{2}kA^2
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
Let's calculate the energy of the wave using two different amplitudes given in the problem:
E_1=\frac{1}{2}k(1)^2=\frac{1}{2}k\\E_2=\frac{1}{2}k(2)^2=\frac{4}{2}k=2k\\
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.


6 0
3 years ago
Read 2 more answers
a 3520 kg truck moving north at 18.5 m/s makes an INELASTIC collision with an 1480 kg car moving east after colliding they have
anyanavicka [17]

Answer:

Explanation:

An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is

p = mv and

p = (3520 + 1480)(13.6) so

p = 68000. Final momentum. The equation for this is a take-off of Pythagorean's Theorem and the one used to find the final magnitude of a resultant vector when you first began your vector math in physics. The equation is

p_f=\sqrt{(p_{truck})^2+(p_{car})^2} which, in words, is

the final momentum after the collision is equal to the square root of the truck's momentum squared plus the car's momentum squared. Filling in:

68000=\sqrt{(65100)^2+(1480v)^2} and

(68000)^2=(65100)^2+(1480v)^2 and

4624000000=4238010000+2190400v^2  and

385990000=2190400v^2 and

176.2189554=v^2 so

v = 13.3 m/s at 72.6°

6 0
2 years ago
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