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Jet001 [13]
3 years ago
9

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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In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o
ioda

Answer:

Explanation:

We shall apply law of conservation of  momentum in space to know the velocity of combination after the impact

m₁v₁ = m₂v₂

.1 x 4 = ( 1 + .1 ) v₂

v₂ = .3636 m /s

1  )  

Kinetic energy of the combination

= 1/2 x 1.1 x ( .3636)²

= 7.3 x 10⁻² J

2 )

Initial kinetic energy of the system

= 1/2 x 0.1 x 4²

= 0.8 J

Final  kinetic energy of the system = 7.3 x 10⁻²

Loss of energy = .8 - .073

= .727 J

This energy was converted into internal energy of the system .

3 )

increase in entropy = dQ / T

Here dQ = .727 J

T  = 300 ( Constant )

dQ / T = 2.42 X 10⁻³ J/K

6 0
3 years ago
Can someone please help meee!!
Paraphin [41]

The answer is 9800 J/S

5 0
3 years ago
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
3 years ago
You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position
Alecsey [184]

Answer:

positive, positive

You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is positive and the velocity of the rock is positive

Explanation:

Given that the ground is defined as the origin.

The position of the rock is positive since the rock is thrown upward, the position also increases with time until it reaches the maximum height. Also, since the rock is thrown upward with the ground as the origin, the velocity of the rock is positive but the velocity reduces with time (change in height per unit time as the rock moves up is positive)

3 0
3 years ago
A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 5.00s what is the speed of the mail
grandymaker [24]
The acceleration of gravity (on Earth) is 9.8 m/s² downward.

This means that every falling object gains 9.8 m/s more downward speed
every second that it falls.

In 5 seconds of falling, it gains (5 x 9.8 m/s) = 49 m/s of downward speed.

If it was already descending at 2.0 m/s at the beginning of the 5 sec,
then at the end of the 5 sec it would be descending at

                                 (2 m/s  +  49 m/s)  =  51 m/s .
7 0
3 years ago
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