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3 years ago

Name the three different types of radiation and describe how they are different in their penetrating abilities. Explain.

Physics

2 answers:

Andreas93 [3]3 years ago

8 0

Answer:

The three different type of radiations are:

1) Alpha (α) radiation

2) Beta (β) radiation

3) Gamma ( $\gamma$ ) radiation

Penetrating Ability:

Penetrating Ability of gamma rays is the highest.

That of Beta Rays is higher.

That of Alpha Rays is lowest

Y_Kistochka [10]3 years ago

7 0

Answer:

1. Alpha radiation

2. Beta radiation

3. Gamma radiation

Explanation:

Penetration Ability:

ALPHA: being relative heavier in mass they have less penetration power.They can penetrate thin layer of solid such as Mica paper. They are stopped by an aluminum foil that is 0.1mm thick

2:BETA:being relative lighter in mass their penetrating power is greater than alpha rays

3: GAMMA:their penetrating power is much greater than other two rays....

Hope it will helps you........

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At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl

bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C) the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0

3 years ago

A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced sli

goblinko [34]

Given Information:

Wavelength of the red laser = λr = 632.8 nm

Distance between bright fringes due to red laser = yr = 5 mm

Distance between bright fringes due to laser pointer = yp = 5.14 mm

Required Information:

Wavelength of the laser pointer = λp = ?

Answer:

Wavelength of the laser pointer = λp = ?

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

y = Dλ/d

y/λ = D/d

Where

λ is the wavelength

y is the distance between bright fringes.

d is the double slit separation distance

D is the distance from the slits to the screen

For the red laser,

yr/λr = D/d

For the laser pointer,

yp/λp = D/d

Equating both equations yields,

yr/λr = yp/λp

Re-arrange for λp

λp = yp*λr/yr

λp = (5*632.8)/5.14

λp = 615.56 nm

Therefore, the wavelength of the small laser pointer is 615.56 nm.

3 0

3 years ago

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$