Which has more mass electron or ion?

What is the speed of sound dependent on?

mojhsa [17]

Depends on the elasticity and density of the medium through what it is traveling <span> </span>

5 0

3 years ago

Can anyone tell me how to read a micrometer screw gauge I want very clear instructions.

Natalka [10]

Explanation:

Things you need to know:

Accuracy refers to the maximum error encountered when a particular observation is made.

Error in measurement is normally one-half the magnitude of the smallest scale reading.

Because one has to align one end of the rule or device to the starting point of the measurement, the appropriate error is thus twice that of the smallest scale reading.

Error is usually expressed in at most 1 or 2 significant figures.

Tape

Equipment: It is made up of a long flexible tape and can measure objects or places up to 10 – 50 m in length. It has markings similar to that of the rigid rule. The smallest marking could be as small as 0.1 cm or could be as large as 0.5 cm or even 1 cm.

How to use: The zero-mark of the measuring tape is first aligned flat to one end of the object and the tape is stretched taut to the other end, the reading is taken where the other end of the object meets the tape.

Ruler

Equipment: It is made up of a long rigid piece of wood or steel and can measure objects up to 100 cm in length. The smallest marking is usually 0.1 cm.

How to use: The zero-end of the rule is first aligned flat with one end of the object and the reading is taken where the other end of the object meets the rule.

Vernier Caliper

Equipment: It is made up of a main scale and a vernier scale and can usually measure objects up to 15 cm in length. The smallest marking is usually 0.1 cm on the main scale.

It has:

a pair of external jaws to measure external diameters

a pair of internal jaws to measure internal diameters

a long rod to measure depths

How to use: The jaws are first closed to find any zero errors. The jaws are then opened to fit the object firmly and the reading is then taken.

Micrometer Screw Gauge

Equipment: It is made up of a main scale and a thimble scale and can measure objects up to 5 cm in length. The smallest marking is usually 1 mm on the main scale (sleeve) and 0.01 mm on the thimble scale (thimble). The thimble has a total of 50 markings representing 0.50 mm.

It has:

an anvil and a spindle to hold the object

a ratchet on the thimble for accurate tightening (prevent over-tightening)

How to use: The spindle is first closed on the anvil to find any zero errors ( use the ratchet for careful tightening). The spindle is then opened to fit the object firmly (use the ratchet for careful tightening) and the reading is then taken.

5 0

3 years ago

A grandfather clock is controlled by a swinging brass pendulum that is 1.2 m long at a temperature of 27°C. (a) What is the leng

stealth61 [152]

Answer:

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

Explanation:

Given;

Initial length L1 = 1.2m

Initial temperature T1 = 27°C

Final temperature T2 = 0.0°C

Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C

The change i length ∆L;

∆L = L2 - L1

L2 = L1 + ∆L ...........1

∆L = xL1(∆T)

∆L = xL1(T2 - T1) ......2

Substituting the given values into equation 2;

∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)

∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)

∆L = -6.156 × 10^-4 m

From equation 1;

L2 = L1 + ∆L

Substituting the values;

L2 = 1.2 m + (- 6.156 × 10^-4 m)

L2 = 1.2 m - 6.156 × 10^-4 m

L2 = 1.1993844 m

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

3 0

3 years ago

In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s

Yanka [14]

"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) $v_{i}$ = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd equation of motion

$h = v_{y_{f}}t + \frac{1}{2}gt^{2}$

-20 = $v_{y_{f}}$ (4) + 0.5(-9.8)(4)²

$v_{y_{f}}$ (4) = 58.4

$v_{y_{f}}$ = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

$v_{f}$ = $v_{y_{f}}$ / sinθ

$v_{f}$ = 14.6 / sin 60

$v_{f}$ = 16.86 m/s

$v_{x_{f}}$ = $v_{f}$ cosθ

$v_{x_{f}}$ = 16.86 cos 60

$v_{x_{f}}$ = 8.43 m/s

To calculate the horizontal distance

d = $v_{y_{f}}$ t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

$v_{x_{f}}$ = $v_{x_{i}}$

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = ( $v_{y_{f}}$ - $v_{y_{i}}$ ) / t

9.8 = 14.6 - $v_{y_{i}}$ ) / 4

$v_{y_{i}}$ = 24.6 m/s

$v_{i}$ = $\sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }$

$v_{i}$ = $\sqrt{8.43^{2}+24.6^{2}}$

$v_{i}$ = 26 m/s

c)

cos β = $v_{x_{i}}$ / $v_{i}$

β = cos⁻¹ (8.43 / 26)

β = 71.08°

3 0

3 years ago

A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2

Pepsi [2]

Work needed = 23,520 J

<h3> Further explanation </h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object

Work is the product of force with the displacement of objects.

Can be formulated

W = F x d

W = Work, J, Nm

F = Force, N

d = distance, m

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

8 0

3 years ago