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sergejj [24]
3 years ago
6

The three-toed sloth is the slowest moving land mammal. On the ground, the sloth moves at an average speed of 0.037 m/s, conside

rably slower m/s? than the giant tortoise, which walks at 0.076 m/s. – After 12 minutes of walking, how much further would the tortoise have gone relative to the sloth
Physics
1 answer:
Georgia [21]3 years ago
7 0

Answer:

tortoise distance w.r.t sloth is 28.08 m further

Explanation:

given data

average speed v1 = 0.037 m/s

walking speed v2 = 0.076 m/s

time t = 12 min = 720 seconds

to find out

how much tortoise have gone wrt sloth

solution

we find here first tortoise walk that is

distance d1 = v2 × t

distance d1 =  0.076  × 720

distance d1 = 54.72 m

and sloth walk distance

distance d2 = v1 × t

distance d2 = 0.037 × 720

distance d2 = 26.64 m

and so

tortoise distance w.r.t sloth = d1 - d2

tortoise distance w.r.t sloth = 54.72 - 26.64 = 28.08

tortoise distance w.r.t sloth is 28.08 m further

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7 0
3 years ago
Read 2 more answers
A bird flies 3.7 meters in 46 seconds, what is its speed?
victus00 [196]

Answer:

Speed is 0.08 m/s.

Explanation:

Given the distance that the bird flies = 3.7 meters

The time is taken by the bird to fly the 3.7 meters = 46 seconds  

We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.  

Below is the formula to find the speed.

Speed = Distance / Time

Now insert the given value in the formula.

Speed = 3.7 / 46 = 0.08 m/s

8 0
3 years ago
A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds
Gwar [14]

Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\

Now replacing the values we have:

a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]

3 0
3 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

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(b)

s_m = 23 \Delta t + 4 \Delta t^2

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3 0
3 years ago
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