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3 years ago

The instrument which is commonly used to measure the intensity of radioactivity is called a

Physics

2 answers:

tatyana61 [14]3 years ago

4 0

The instrument is a Geiger counter and is used to measure radioactive level around people's bodies.

kifflom [539]3 years ago

3 0

The instrument which is commonly used to measure the intensity of radioactivity is called a geiger counter

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A book is on the table. If the weight of the book is 25 newtons, what is the magnitude and direction of the normal force?

Andrej [43]

I'm not sure about the magnitude, but the direction of the normal force is upward.

4 0

3 years ago

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A student whirls a rubber stopper attached to a string in a horizontal circle above her head. If the radius of the stopper in ci

alexandr402 [8]

Answer:

$v = 12.57 m/s$

Explanation:

As we know that the radius of the circular motion is given as

$R = 50.00 cm$

time period of the motion is given as

$T = 0.2500 s$

now we know that it is moving with uniform speed

so it is given as

$v = \frac{2\pi R}{T}$

now plug in all data

$v = \frac{2\pi(0.50)}{0.25}$

$v = 12.57 m/s$

3 0

3 years ago

A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

3 0

3 years ago

Read 2 more answers

Why is a camera lens round but the pictures come out square

sergey [27]

It is round or say spherical to widen the range of photography and it is designed to focus the image as the ray are coming from infinity so !!

I am not an expert of camera but ya it is the main theme !!

6 0

3 years ago

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When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

Masteriza [31]

Answer:

b) True. the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

fr - w = ma

Where the friction force has some form of type.

fr = G v + H v²

Let's replace

(G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

fr - w = 0

fr = mg

(G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

G v = mg

v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0

3 years ago