A chipmunk with a mass of 2 kg sits on a branch that is 10 m off the ground.

A motorcyclist accelerates from rest to 10 mi/hr. what is the change in velocity

gulaghasi [49]

The change in velocity is 10 mi/h (4.47 m/s)

Explanation:

The change in velocity of the motorcyclist is given by

$\Delta v = v-u$

where

v is the final velocity

u is the initial velocity

In this problem, we have

u = 0 (the motorbike starts from rest)

v = 10 mi/h

Therefore, the change in velocity is

$\Delta v = 10 -0 = 10 mi/h$

And keeping in mind that

1 mile = 1609 m

1 h = 3600 s

We can convert it into m/s:

$\Delta v = 10 \frac{mi}{h} \cdot \frac{1609 m/mi}{3600 s/h}=4.47 m/s$

Learn more about velocity:

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6 0

3 years ago

Which component of weight is cause for S.H.M of bob simple pendulum

jeka57 [31]

Explanation:

The Simple Pendulum. A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length L with negligible mass ((Figure)). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string.

hope it helps you

6 0

2 years ago

A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t

Readme [11.4K]

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)

5 0

3 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:

a)

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).

b)

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,

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

c)

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,

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

d)

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,

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

e)

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

Which describes the best approach when conducting a scientific experiment

IgorC [24]

When conducting and experiment you want to have a notebook and something to write down notes with so you can keep everything organized and proper, and to not miss anything in the experiment. Also you want to have everything in order of the way it should be in.

I hope you found this helpful!

5 0

3 years ago