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o-na [289]
2 years ago
6

Natalia is studying a wave produced in her magnetics lab. This wave can move through the empty space in a vacuum and carries a l

ot of energy. What wave is Natalia most likely studying?
Physics
2 answers:
astra-53 [7]2 years ago
8 0

Answer:

A gamma ray

Explanation:

Someone else said this...

jarptica [38.1K]2 years ago
3 0

Answer:

a gammawave

Explanation

You might be interested in
Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc
lubasha [3.4K]

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

8 0
2 years ago
Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
1 year ago
What is Yusef most likely finding?
Ratling [72]

Explanation:

yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .

4 0
1 year ago
A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel ro
Gelneren [198K]

Answer:

59cm

Explanation:

angular velocity = 0.8 rad/s

linear velocity = angular velocity * radius

                        =0.8rad/s * 5m

                        = 4 m/s

wavelength = (V + U)/F

where,

V is the velocity of the wave

U is the velocity of the source

F is the frequency of the source.

wavelength = (350 m/s + 4 m/s ) / 600 Hz

Wavelength = 0.59m or 59 cm

4 0
2 years ago
A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle
s344n2d4d5 [400]

Answer:

0.24 kgm²

Explanation:

L = length of the bat = 81.3 cm = 0.813 m

m  = mass of the bat = 0.96 kg

d  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

T  = Period of oscillation = 1.35 sec

I = moment of inertia of the bat

Period of oscillation is given as

T = 2\pi \sqrt{\frac{I}{mgd}}

1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}

I = 0.24 kgm²

6 0
2 years ago
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