All categories

3 years ago

What did Dr. Kidwell's fighting so the birds eating a box on different tree trunks

Physics

1 answer:

Gelneren [198K]3 years ago

3 0

I think it's D(Leaves).

You might be interested in

How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the

GaryK [48]

Answer:

Explanation:

The heat required to change the temperature of steam from 125.5 °C to 100 °C is:

$Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J$

The heat required to change the steam at 100°C to water at 100°C is;

$Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J$

The heat required to change the temperature from 100°C to 0°C is

$Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J$

The heat required to change the water at 0°C to ice at 0°C is:

$Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J$

The heat required to change the temperature of ice from 0°Cto -19.5°C is:

$Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C) \\ \\ Q_5 = 7132.125 \ J$

The total heat required to change the steam into ice is:

$Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J$

The time taken to convert steam from 125 °C to 100°C is:

$t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W} = 8.12 \ s$

The time taken to convert steam at 100°C to water at 100°C is:

$t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s$

The time taken to convert water to 100° C to 0° C is:

$t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s$

The time taken to convert water at 0° to ice at 0° C is :

$t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08 \ s$

The time taken to convert ice from 0° C to -19.5° C is:

$t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55 \ s$

5 0

3 years ago

A 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. At one moment it is moving 5.0 m/s. A little l

Y_Kistochka [10]

The change in kinetic energy of the car is equivalent to the change in its potential energy. Thus:
K.E = P.E
1/2 x mΔv² = mgΔh
h = (8.2² - 5²) / 2(9.81)
h = 2.15 meters

4 0

3 years ago

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0

2 years ago

A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc

OLga [1]

Answer:

5 N

Explanation:

The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N

7 0

2 years ago

Read 2 more answers

Please help me, I beg you

MA_775_DIABLO [31]

Answer:

40g

Explanation:

Solubility of Copper sulfate at 90°=60g

Solubility of potassium bromide at 90°=100g

100g-60g=40g

8 0

2 years ago