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3 years ago

Why aren't descriptive investigations repeatable ?

Physics

2 answers:

Anna [14]3 years ago

7 0

Answer:

Descriptive investigations are not repeatable because they are based only on observations made at a single point in time. The results may vary at a different time. In addition, descriptive investigations do not contain variables that may indicate cause-and-effect relationships

Explanation: your welcome ;))

Ann [662]3 years ago

4 0

Because the information cant be out of the investigation

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Dolphins and other sea creatures can leap to great heights by swimming straight up and exiting the water at a high speed. A 200

scoundrel [369]

Answer:

i don't know sorry

Explanation:

so if a pie is baked

5 0

3 years ago

A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.

ANTONII [103]

Answer: The number of photons are $3.7\times 10^8$

Explanation:

We are given:

Wavelength of microwave = $1.22\times 10^8nm=0.122m$ (Conversion factor: $1m=10^9nm$ )

To calculate the energy of one photon, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = 0.122 m

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

4 0

4 years ago

Which of the following statements must be true if the net force on an object is zero? Choose all that apply.

Black_prince [1.1K]

Answer:

a) The object must have constant velocity.

d) The object must have zero acceleration.

Explanation:

We can solve the problem by using Newton's second law, which states that the net force acting on an object is equal to the product between mass and acceleration:

$F = ma$

where

F is the net force

m is the mass of the object

a is the acceleration

In this problem, the net force on the object is zero:

F = 0

This means that the acceleration of the object is also zero, according to the previous equation:

a = 0

So statement (d) is correct. Moreover, acceleration is defined as the rate of change of velocity:

$a=\frac{\Delta v}{\Delta t}$

Which means that $\Delta v=0$ , so the velocity is constant. Therefore, statement (a) is also correct. The other two statements are false because:

b)The object must be at rest. --> false, the object can be moving at constant velocity, different from zero

c)The object must be at the origin. --> false, since the object can be in motion

5 0

3 years ago

The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The t

Gemiola [76]

Answer:

$\omega=7.16*10^{-13}\frac{rad}{s}$

Explanation:

The angular speed is given by:

$\omega=\frac{v}{r}$

Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to $\frac{m}{s}$ :

$1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}$

Now, we calculate the angular speed:

$\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}$

8 0

4 years ago

A crane lifts up two boxes. Which free body diagram shows the forces acting on block A?

Temka [501]

Answer:

The 3rd graph

Explanation:

A free body diagram is a diagram which shows all the forces acting on an object.

The problem asks us to find the free body diagram of block A, so we must find all the forces acting on block A.

We have 3 forces acting on block A in total:

- The force of gravity (its weight), which pushes the block downward (in the diagram, it is the force represented with $F_{gA}$

- The tension in the rope 1, which pulls block A upwards: this force is represented with $F_{T1}$

- The tension in the rope 2, due to the weight of block 2, which pulls block A downwards: this force is represented with $F_{T2}$

Based on the direction of these 3 forces, the correct diagram is the 3rd one.

7 0

3 years ago

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