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zhuklara [117]
2 years ago
10

Which type of radio active decays​

Physics
1 answer:
stepan [7]2 years ago
3 0
Negative beta decay. So probably 4.
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Calculate the minimum frequency of ultrasound (in Hz) that will allow you to see details as small as 0.193 mm in human tissue. (
STatiana [176]

Answer:

f = 7.97 x 10⁶ Hz = 7.97 MHz

Explanation:

The speed of a wave is given by the following formula:

v = f\lambda

where,

v = speed of the ultrasound wave through human tissue = 1540 m/s

f = frequency of ultrasound wave required = ?

λ = wavelength of ultrasound waves = smallest detail required = 0.193 mm

λ = 0.193 mm = 1.93 x 10⁻⁴ m

Therefore,

1540\ m/s = f(1.93\ x\ 10^{-4}\ m)\\f = \frac{1540\ m/s}{1.93\ x\ 10^{-4}\ m}

<u>f = 7.97 x 10⁶ Hz = 7.97 MHz</u>

8 0
3 years ago
How can I rewrite the equation a - b = d using addition?
Flauer [41]

Explanation:

A=b+d that is the way to rewrite the equation

5 0
2 years ago
Which of the following statements are true of thermal energy and kinetic
Elis [28]

Answer:

a

all the molecules or atoms in motion has kinetic energy

4 0
1 year ago
a small asteroid of mass 125 kg is orbiting a planet that has a mass of 3.52x 10^13 what is the radial distance between the aste
asambeis [7]

Answer:

r = 2.031 x 10⁶ m = 2031 km

Explanation:

In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:

Centripetal Force = Gravitational Force

mv²/r = GmM/r²

v² = GM/r

r = GM/v²

where,

r = radial distance = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Planet = 3.52 x 10¹³ kg

v = tangential speed = 0.034 m/s

Therefore,

r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²

<u>r = 2.031 x 10⁶ m = 2031 km</u>

7 0
3 years ago
Calculate the force of gravity on the 0.60- kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth
nignag [31]
The gravitational force between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the first object has a mass of m_1=0.60 kg, while the second "object" is the Earth, with mass m_2=5.97 \cdot 10^{24}kg. The distance of the object from the Earth's center is r=1.3 \cdot 10^7 m; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
F=G \frac{m_1m_2}{r^2}=(6.67\cdot 10^{-11}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=  1.41 N
5 0
3 years ago
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