How much work would be required to lift a 12.0 kg mass up onto a table 1.15 meters high

A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e

IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0

3 years ago

A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as

Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

$T= 3600\ sec$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega=\dfrac{2\pi}{3600}$

$\omega=0.001745\ rad/s$

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

$a=r\omega^2$

Put the value into the formula

$a=0.55\times(0.001745)^2$

$a=1.675\times10^{-6}\ m/s^2$

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

3 0

2 years ago

Wire 1 has a resistance R1. Wire 2 is made of the same material, with length 1/2 as long and diameter 1/4. What is the resistanc

hodyreva [135]

-- reduce the length of a wire to 1/2 . . . cut the resistance in half

-- reduce the diameter to 1/4 . . . reduce the cross-section area by (1/4²) . . . increase the resistance by 16x .

-- R2 = (R1) · (1/2) · (16) = 8 · R1

5 0

2 years ago

A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini

Volgvan

Answer:

The horizontal range will be $2.55\times 10^5m$

Explanation:

We have given initial speed of the shell u = $1.6\times 10^3m/sec$

Angle of projection = 51°

Acceleration due to gravity $g=9.8m/sec^2$

We have to find maximum range

Horizontal range in projectile motion is given by

$R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m$

So the horizontal range will be $2.55\times 10^5m$

6 0

2 years ago

Zinc can not be changed into a simpler substance by a chemical or physical process. Therefore, zinc is classified as:

bonufazy [111]

The correct answer I think is a compound

7 0

2 years ago