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Norma-Jean [14]
3 years ago
10

Suppose you have a large boulder in your yard that you’d like to move. How could you break it into smaller pieces without using

a sledgehammer?
Physics
2 answers:
Scrat [10]3 years ago
8 0

Is the question that is trying to be answered here how to move the boulder, or how to break up a boulder? If the purpose of the question is how to move the boulder from A to B, then you roll it.

laiz [17]3 years ago
8 0

-- Build a long, skinny campfire all around the rock, as close as possible to the rock, so the hot coals are right around the underside of the rock, and the flames lick up and around it.

-- Let it burn there for a few hours, so that a lot of the outside of the rock gets really hot.

-- Throw a big bunch of cold water on top of the rock ... enough so that it runs down the whole outside of the rock.

The thermal stresses that suddenly propagate through the rock, on the surface and for some distance into it, crack a layer of some thickness into several pieces, and the pieces fall off onto the ground.

If the cracking and crumbling doesn't extend far enough in, then just pick up the pieces of the layer that DID crumble off, and then go through the  whole same process again, with the inside smaller rock that remains.

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How are heat and light waves produced on and in the sun?
enyata [817]
All of the electromagnetic energy radiated from the sun (and from
other stars) is the product of nuclear fusion in its core.
8 0
3 years ago
Read 2 more answers
Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0
3 years ago
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
Light travels in a straight line at a constant speed of 3,0 x 108 m/s for 4,1
zepelin [54]

Answer:

As the velocity of light is constant so the acceleration of the light is equal to zero.

a= dv/dt

Explanation:

4 0
3 years ago
If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp
Mice21 [21]

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

m_1(v_f - v_i) = m_2(\Delta v)

given that

m_1 = 5 ounce

m_2 = 32 ounce

For ball the change in speed will be

v_f - v_i = (120 - 77)mph

now from above equation

5\times (120 - 77) = 32 \times \Delta v

\Delta v = 6.72 mph

so speed of bat will decrease by 6.72 mph

3 0
3 years ago
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