Help me plz i'll mark brainliest

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 = 4.2 m/s

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 = 9.9 m

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 = 49.1 m

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

calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutro

Orlov [11]

To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

$\rho = \frac{m}{V}$

Where

m = Mass (Neutron at this case)

V = Volume

The mass of the neutron star is 1.4times to that of the mass of the sun

The volume of a sphere is determined by the equation

$V = \frac{4}{3}\pi R^3$

Replacing at the equation we have that

$\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}$

$\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}$

$\rho = 5.75*10^{17}kg/m^3$

Therefore the density of a neutron star is $5.75*10^{17}kg/m^3$

4 0

3 years ago

Find the average velocity of the car during the time last three second interval. Answer in units of m/s.

Pani-rosa [81]

Answer:

Average Velocity:

When a body covers some displacement moving from point A to B, in a given time,t. Then dividing the total distance covered by the time taken to complete the displacement is called as the average velocity of that car.

We can calculate the average velocity of car by subtracting the initial displacement of a body,s₁ from the final displacement position of the body, s₂. And then dividing by the initial time,t₁ been subtracted from the final time,t₂.

Δv=x₂-x₁/t₂-t₁,

We will get the required results, now if we want to obtain the results in the final three seconds then, we can have the following explanation to that as given.

Explanation:

Lets say, that we have the car moving for 10 seconds and then it stops at the eleventh second of the trip on final position, of the displacement. Then we divide the whole trips into small segments, as we will have the different velocity at each interval but, the velocity is unknown at the last 3 seconds is unknown=?.

Total time taken, Δt= 11 seconds,
Then lets say the velocity at the 8th second is , v=25 m/s,then
Δv=x₂-(25)/11-0,⇒Δv=x₂-25/11,

3 0

3 years ago

The 20 g bullet is traveling at 400 m/s when it becomes embedded in the 2 -kg stationary block. Determine the distance the block

Marina CMI [18]

Answer:

The block+bullet system moves 4 m before being stopped by the frictional force.

Explanation:

Using the law of conservation of llinear momentum and the work energy theorem, we can obtain this.

According to Newton's second law of motion

Momentum before collision = Momentum after collision

Momentum before collision = (0.02×400) + 0 (stationary block)

Momentum before collision = 8 kgm/s

Momentum after collision = (2+0.02)v

8 = 2.02v

v = 3.96 m/s.

According to the work-energy theorem,

The kinetic energy of the block+bullet system = work done by Friction to stop the motion of the block+bullet system

Kinetic energy = (1/2)(2.02)(3.96²) = 15.84 J

Work done by the frictional force = F × (distance moved by the force)

F = μmg = 0.2(2.02)(9.8) = 3.96 N

3.96d = 15.84

d = (15.84/3.96) = 4 m

5 0

3 years ago

If a raindrop is 0.05 cm3 in volume, what would the volume of an avogadro number of raindrops be?

zhenek [66]

If a raindrop is 0.05 cm^3 in volume, then the volume of an Avogadro number of raindrops be 3.011 x 10^22 cm^3. To calculate this value, we simply multiply the volume given to Avogadro's number which represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23.

4 0

3 years ago