All categories

3 years ago

Question 4 of 10

Physics

1 answer:

ch4aika [34]3 years ago

4 0

Habitat fragmentation is a cost of urban development.

Option: A

Explanation:

Though from the view point or perspective of up gradation and development urban development is much needed but in cost of habitat fragmentation which feels very bitter. As habitat fragmentation leads to the loss of habitat, disruption of ecological cycle and environmental equilibrium.

Actually in the name of urban development we the human use our bread giver environment in a wrong way which causes natural disasters in long run. Animals become endangered , vulnerable and extinct with passage of time. Because they forced to enter into human settlements.

You might be interested in

You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion

Nina [5.8K]

Answer:

v (minimum speed) = 2.90 m/sec.

$\\ \\ maximum speed (v)= 6.57 m/sec.\\$

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?

Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

$m*g + T = m*v^2/R$

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

$v = \sqrt{g*R}$

$v = \sqrt{9.81*0.86}$

v (minimum speed) = 2.90 m/sec.

b) what is the maximum speed the rock can have so that the string does not break?

Here the force balance at bottom of circle is represented by the illustration:

$T = m*g + m*v^2/R$

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass m = 0.75 kg

∴

$45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\$

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because at the lowest point; the tension in string will be maximum.

4 0

3 years ago

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the

ikadub [295]

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $\frac{velocity}{2 *length}$

velocity = $\sqrt{\frac{tension}{mass per unit length} }$

mass per unit length= $\frac{3.5}{1000*1.22}$ =0.00427 $\frac{kg}{m}$

Now calculating velocity v= $\sqrt{\frac{255}{0.00427} }$

=244.3 $\frac{m}{sec}$

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = $\frac{244.3}{2 *0.7}$ =174.5 hz

6 0

3 years ago

A submarine travels 25 km/h north for 3.2 hours. What is its displacement?

Vika [28.1K]

displ = velocity x time

25 x 3.2 = 75+5 km north.

7 0

3 years ago

Read 2 more answers

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0

3 years ago

When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere

Sergeeva-Olga [200]

Lower stratosphere, this is to avoid turbulence

6 0

3 years ago

Read 2 more answers