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3 years ago

8

The blade of a metal cutter is shorter than that of the scissors used by tailor.why

1 answer:

garik1379 [7]3 years ago

4 0

Answer:

It depends on the amount of pressure needed to cut your material. Metal is a tougher material to cut though, so the blades must be shorter to create more pressure to break through the metal. Cloth on the other hand is easier to cut through so the blades can be longer in order to cut more in each snip.

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A 2.16 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const

pochemuha

Answer:

0.15694 m

Explanation:

m = Mass of block = $2.16\times 10^{-2}\ kg$

v = Velocity of block = 11.2 m/s

k = Spring constant = 110 N/m

Here the kinetic energy of the fall and spring are conserved

$\frac{1}{2}mv^2=\frac{1}{2}kA^2\\\Rightarrow mv^2=kA^2\\\Rightarrow A=\sqrt{\frac{mv^2}{k}}\\\Rightarrow A=\sqrt{\frac{2.16\times 10^{-2}\times 11.2^2}{110}}\\\Rightarrow A=0.15694\ m$

The amplitude of the resulting simple harmonic motion is 0.15694 m

3 0

3 years ago

under what circumstances can the average velocity of a moving object be zero when its average speed is 50 km/hr ?

german

If the displacement = zero the average velocity will be zero. displacement = zero when the object move and return back to its starting position

5 0

3 years ago

A tennis ball is dropped from 1.62 m above the

natita [175]

Answer:

-5.63 m/s

Explanation:

Given:

y₀ = 1.62 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.62 m)

v = -5.63 m/s

8 0

3 years ago

Calculate the acceleration due to gravity on the moon. the moon's radius is 1.74×106m and its mass is 7.35×1022 kg.

Kaylis [27]

184.44

7,511.7 that s the answer

7 0

3 years ago

A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse

bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, $f=4.2\times 10^{14}\ Hz$

Time, $t=3.2\times 10^{-11}\ s$

(a) Let $\lambda$ is the wavelength of this light. It can be calculated as :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}$

$\lambda=7.14\times 10^{-7}\ m$

or

$\lambda=714\ nm$

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

$n=f\times t$

$n=4.2\times 10^{14}\times 3.2\times 10^{-11}$

n = 13440

Hence, this is the required solution.

8 0

2 years ago