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Ivahew [28]
3 years ago
5

Which apparatus Is used to determine the volume of a small block of unknown material

Physics
1 answer:
Aleks04 [339]3 years ago
8 0

Options are;

A: measuring cylinder, metre rule

B: measuring cylinder, stopwatch

C: metre rule, balance

D: metre rule, stopwatch

Answer:

A: measuring cylinder, metre rule

Explanation:

Usually, when measuring volume of a small material, we make use of a piece of laboratory apparatus known as measuring cylinder which is also known as graduated cylinder.

Since it is a graduated cylinder, it means we will make use of metre rule.

Thus, the correct option is option A.

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A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:
AfilCa [17]

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

F = mv/t

Given the following parameters

Force F = 8.0N

mass m = 16kg

time t = 4.0s

Required

speed v

Substitute the given parameters into the formula

v = Ft/m

v = 8 * 6/16

v = 48/16

v = 3m/s

Hence the change in speed of this object is 3m/s

Learn more here: brainly.com/question/19072061

8 0
3 years ago
Does this look like a good circuit diagram?
AnnyKZ [126]

It's hard to tell what's going on down there in the corner with the resistor and the ammeter. There seems to be as many as 3 or 4 wires in and out of the ammeter, which would be wrong. A real ammeter only has two ... one in and one out. (Same for a resistor.)

It's hard to say whether this circuit works, until we can clearly understand how everything is hooked up in that corner of the drawing.

6 0
3 years ago
An airplane undergoes the following displacements: First, it flies 40 km in a direction 30° east of north. Next, it flies 56 km
goldfiish [28.3K]

Answer:

Distance from start point is 72.5km

Explanation:

The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:

x1=d1*cos(60), y1=d1*sin(60)\\\\ x2=x1 , y2=y1-d2\\\\ x3=x2-d3*cos(30) , y3=y2+d3*sin(30)

Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:

d=\sqrt{x3^{2} +y3^{2} }

Replacing the given values in the equations, the distance is calculated.

4 0
3 years ago
Calculate 2 x 10^-3cm ÷ 2.5 x 10^4cm
Mamont248 [21]
Here is your answer:

First find the notations:

2×10^-3
=0002
And...
2.5×10^4=25000

Then divide:

0002÷25000=8E-9

Your answer:
=8 x 10-8
3 0
3 years ago
An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.
rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = \frac{1}{2}m_{p}v²

v = √( 2K / m_{p} )

lets relate the cross-sectional area A of the beam to its diameter D;

A = \frac{1}{4}πD²

now, we substitute for v and A

n = I / \frac{1}{4}πeD² ×√( 2K / m_{p} )

n = 4I/π eD² × √(m_{p} / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n =  1.98695 × 10¹⁸ × 1.6157967  × 10⁻⁵

n = 3.2 × 10¹³ m⁻³  

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0
2 years ago
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