Help please!!!

Which of the following would not be used to detect radioactivity?

igor_vitrenko [27]

The cloud chamber, the Geiger counter, and a piece of
photographic film can all reveal the presence of radioactivity.
Batteries don't.

8 0

4 years ago

Read 2 more answers

For the electric circuit given below, calculate: 3 (i) the equivalent resistance of the circuit, (ii) the total current drawn fr

Daniel [21]

Answer:

1)31/3Ω

2)18/31A

3)4.06V

Explanation:

According to the diagram and 10 Ω resistors are parallel to each other, parallel resistor can be calculated using the formula below

1/R= 1/R1 + 1/R2

But we know R1= 5 Ω and R2= 10 Ω

1/R= 1/5 + 1/10

R= 10/3 Ω

=3.33 Ω

Then if we follow the given figure, 10/3 Ω and 7Ω are now in series then

Req = 10/3 +7

Req= 31/3 Ω

Therefore, equivalent resistance = 31/3 Ω

According to ohms law we know that V= IR

Then I= V/R

Where I= current

R= resistance

V= voltage

I= 6/(31/3)

I= 18/31A

We can now calculate the voltage accross the resistor which is

V=(18/31)× 7

V=4.06V

Therefore, the voltage accross the 7 ohm resistor is 4.06V

CHECK THE FIQURE AT THE ATTACHMENT

3 0

3 years ago

In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a

katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

brainly.com/question/11422992

#SPJ1

3 0

2 years ago

Seeds are often found on which part of a gymnosperm?<br><br> branch<br> leaf<br> cone<br> stem

inn [45]

Cone i hope this helps

8 0

3 years ago

You slide a chair across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 156 N dir

Luda [366]

Answer:

626.612 J

Explanation:

Work done by friction on the chair is given as

W-ΔEk = W'..................... Equation 1

Where W' = Work done by friction on the chair, W = Work done on the chair by me, Ek = change in Kinetic energy of the chair as a result of the slide.

From the question,

W = FdcosФ.............. Equation 2

ΔEk = 1/2m(v²-u²)................ Equation 3

Where F = Force applied on the chair, d = distance of slide, Ф = angle between the force and the horizontal, m = mass of the chair, v = final velocity of the chair, u = initial velocity of the chair

Substitute equation 2 and equation 3 into equation 1

W' = FdcosФ-1/2m(v²-u²)........................ Equation 4

Given: F = 156, d = 5 m, Ф = 26°, m = 18.8 kg, v = 3.1 m/s, u = 1.3 m/s

Substitute into equation 4

W' = 156×5×cos26°-1/2×18(3.1²-1.3²)

W' = 701.06-74.448

W' = 626.612 J.

Hence the work done by friction = 626.612 J

8 0

3 years ago