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3 years ago

A bike travels at 3.0 m/s, and then accelerates to a speed of

Physics

1 answer:

suter [353]3 years ago

7 0

Answer:

2.2m/s²

Explanation

Initial velocity of bike = 3m/s

Final velocity of bike = 8.5m/s

Total time = 2.5 seconds

Acceleration = (Final velocity - initial velocity) / time

= 8.5 - 3 / 2.5

= 5.5/2.5 => 2.2 m/s²

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An electric motor rotates 60 times per second if the alternating current source is 60 Hz. How many times will an electric motor

valentina_108 [34]

Answer:

180,000

Explanation:

Frequency is a quantity that is measured in Hertz [Hz] and it represents the number of rotations per second.

A motor with a frequency of 50 Hz will rotate 50 times per second.

Since we don't want to know how many times it rotates per second, but per hour. The first step is to find how many seconds there are in an hour and then multiply that amount by 50.

Seconds in an hour:

there are 60 seconds per minute, and 60 minutes per hour, thus there are

60*60 = <u>3,600 seconds in an hour</u>

We know that the motor will rotate 50 times per second so to find the number of rotations in 1 hour = 3,600 seconds we multiply:

50*3,600 = 180,000 rotations

8 0

3 years ago

Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing of N 42º E from the western sta

olchik [2.2K]

Answer:

Explanation:

AB = 110 miles

Let the distance of the western station from fire is d.

As according to the diagram, use Sine law

$\frac{d}{Sin 15}=\frac{110}{Sin 133}$

d = 110 x 0.2588 / 0.73

d = 39 miles

8 0

3 years ago

What is the magnitude of the kinetic frictional force

Effectus [21]

The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps

4 0

3 years ago

A body starts from rest and travels ‘s’ m in 2nd second, than acceleration is.

Lana71 [14]

Answer:

Explanation:

3 0

2 years ago

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

3 years ago