Answer questions 1 & 2

A clothes dryer uses about 5 amps of current from a 240 volt line. How much power does it use?

yan [13]

PVI, so the power used would be 240*5= 1200 Watts.

6 0

3 years ago

How do you find density of a regular solid?

Effectus [21]

Put the object or material on a scale to figure out its mass. 3. Divide the mass by the volume to figure out the density (p = m / v). You may also need to know how to calculate the volume of a solid so use the formula

3 0

3 years ago

HELP ASAPPP which statement explains the first law of thermodynamics a)energy is created and destroyed or b) energy is transform

Aliun [14]

B I believe... :) good luck!

7 0

4 years ago

Read 2 more answers

A water bug is suspended on the surface of a pond by surface tension (water does not wet the legs). The bug has six leg, and eac

Crazy boy [7]

Answer:

m = 2.2 x 10⁻⁴ kg = 0.22 g

Explanation:

The surface tension of water is 0.072 N/m. So in order for the bug to avoid sinking, its weight per unit length of contact must be no more than the surface tension of water. Therefore,

$Weight\ of bug\ per\ unit\ length = Surface\ Tension\ of\ Water\\\frac{mg}{L} = Surface\ Tension\ of Water\\m = \frac{(Surface\ Tension\ of\ Water)(L)}{g}$

where,

m = mass of bug = ?

g = acceleration due to gravity = 9.81 m/s²

L = Contact length = (contact length of each leg)(No. of Legs) = (5 mm)(6)

L = 30 mm = 0.03 m

Therefore,

$m = \frac{(0.072\ N/m)(0.03\ m)}{9.81\ m/s^{2}} \\$

m = 2.2 x 10⁻⁴ kg = 0.22 g

8 0

3 years ago

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

8 0

3 years ago