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4 years ago

11

____________ occurs when an energy source transfers heat directly to another object through space; an example would be an object

becoming warm by sitting in the sunshine.
a. conduction
c. convection
b. radiation
d. heat

2 answers:

larisa86 [58]4 years ago

8 0

Answer:

B or the second option (Radiation) for Edg 2020

Explanation:

timurjin [86]4 years ago

6 0

The answer is Radiation

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What properties does a loud, shrill whistle have? a.) high amplitude, high frequency

marysya [2.9K]

Answer:

a.) high amplitude, high frequency

Explanation:

Frequency and amplitude are properties of sound. Varying these properties changes how people perceive sound.

While hearing sound of a particular frequency we call it pitch i.e., the perception of a frequency of sound.

High pitch means high frequency and high frequency is perceived to have a shrill sound.

The loudness of a sound is measured by the intensity of sound i.e., the energy the sound possesses per unit area. As the amplitude increases the intensity increases. So, a loud sound will have higher density.

Hence, the loud shrill whistle will have high frequency and high amplitude.

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3 years ago

Need help with these two questions Please?

Ksivusya [100]

Im too young for physics, sorry!

5 0

3 years ago

A 6 cm long spring extends to 9 cm when a 1 kg load is suspended from it. What would be its length if a 2 kg load were suspended

Ganezh [65]

The answer is D that is the right question

8 0

3 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

Read 2 more answers

Differnces between uniform and non uniform electric fields

Soloha48 [4]

The electric field strength of a uniform electric field is constant throughout the field. A perfectly uniform electric field has no variations in the entire field and is unattainable in the real world. However, two parallel plates can generate a field that resembles a perfectly uniform field with slight variations near the edge of the plates. <span>Electric fields are represented by drawing field lines that represent the direction of the field, as well as the strength of the field. More field lines represents a higher field strength. In a non-uniform electric field, the field lines tend to be curved and are more concentrated near the charges. In a uniform electric field, since the field strength does not vary, the field lines are parallel to each other and equally spaced. Uniform fields are created by setting up a potential difference between two conducting plates placed at a certain distance from one another. The field is considered to be uniform at the center of the plates, but varies close to the edge of the plates. The strength of the field depends on the potential difference applied to the plates and the distance by which they are separated. A higher potential difference or voltage results in a stronger electric field. The greater the distance between the plates, the weaker the field becomes. The electric field is therefore calculated as a ratio of the voltage between the plates to the distance they are separated by.</span>

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3 years ago