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Masja [62]
3 years ago
14

Select the correct answer. The rules of volleyball today have been only slightly changed from the rules when the game was A True

B. False Next​
Physics
1 answer:
Rasek [7]3 years ago
3 0
B bbbbb bbbbbbbbbbbbbb bbbbbbbbbbbbbbbbbbbbb
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A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at  t = 4.7 s will be:

v = v0 + a · t    (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = <u>4.2 m/s</u>

<u />

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = <u>9.9 m</u>

<u />

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x =  v · t

x = 4.2 m/s · 11.7 s = <u>49.1 m</u>

<u />

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t²      (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0
3 years ago
The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t
Margaret [11]
The vertical component is given by the sin of the angle.  So this is 20*sin(53)=15.97m/s
8 0
3 years ago
State two other ways in which evaporation is different from boiling​
Nookie1986 [14]

Answer:

To summarize, evaporation is slower, occurs only from the surface of the liquid, does not produce bubbles, and leads to cooling. Boiling is faster, can occur throughout the liquid, produces lots of bubbles, and does not result in cooling.

Explanation:

3 0
3 years ago
Read 2 more answers
Controlled variables
PilotLPTM [1.2K]

Answer:

A control variable in scientific experimentation is an experimental element which is constant and unchanged throughout the course of the investigation.

Explanation:

7 0
3 years ago
A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an
Oxana [17]

Answer:

Explanation:

Mass of 190kg

Coefficient of static friction is 0.4

Coefficient of kinetic friction 0.36

Horizontal force= 500N

Taking g=9.81m/s^2.

The weight of the body my

W=190×9.81=1863.91N

There is a normal acting on the body which is equal to the weight

N=W=1863.91N

Frictional force(fr) is acting on the body and it is opposite the horizontal force.

The minimum force to be overcome before the object can start to move is Fr = μsN

Fr= μsN. μs=0.4

Fr= 0.4×1863.91

Fr=745.56N.

Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

Then the frictional force at that time is equal to the horizontal force

Therefore

Functional force = 500N

b. Mass of asteroid is

M=2000kg

Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

U=√1.933E9

U=4.39×10^4m/s

Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

R=√3.6E13+10E13+0

R=√13.6E13

R=1.17E7m

R^2=13.6E13m

The mass of the earth is

Me=5.97x10^24 kg

The momentum of the asteroid after time, t=1.5×10^3s

Given that G=6.67x10^-11Nm^2/kg^2

Momentum is

Mv-Mu=Ft

There the new momentum will be

Mv=Ft+Mu

Now we the to find the force the earth exert on the asteroid by using

F=GMMe/R^2

F=6.67E-11 ×2000× 5.97E24 /13.6E13

F=7.964E17/13.6E13

F=5855.88N

The new momentum

Mv= Mu+Ft

Mv= 2000(4.39E4)+5855.88(1.5E3)

Mv=9.66E7kgm/s

The new momentum is 9.66×10^7 Kgm/s

8 0
3 years ago
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