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3 years ago

A 1200N load is to be lifted with 200N effort using a first class lever. At what distance

Physics

1 answer:

kipiarov [429]3 years ago

7 0

Explanation:

Hey there!!

Let's simply work with it.

Here,

load = 1200N

Effort = 200N

Load distance = 15cm

We have,

According to the principle of lever.

L×LD = E×ED.

1200×15 = 200× ED.

18000 = 200ED.

$ed = \frac{18000}{200}$

Therefore, Effort Distance = 90cm.

Hope it helps....

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A certain tuning fork vibrates at a frequency of 215 Hz while each tip of its two prongs has an amplitude of 0.832 mm. (a) What

Marysya12 [62]

Explanation:

It is given that,

Frequency of vibration, f = 215 Hz

Amplitude, A = 0.832 mm

(a) Let T is the period of this motion. It is given by the following relation as :

$T=\dfrac{1}{f}$

$T=\dfrac{1}{215}$

$T=4.65\times 10^{-3}\ s$

(b) Speed of sound in air, v = 343 m/s

It can be given by :

$v=f\times \lambda$

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

$\lambda=\dfrac{343\ m/s}{215\ Hz}$

$\lambda=1.59\ m$

Hence, this is the required solution.

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3 years ago

A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 280.5 meters from the wall and the sound of h

ValentinkaMS [17]

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

t= 4.80 seconds

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3 years ago

A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac

lana66690 [7]

Answer:

charges of the beads is 1.173 × $10^{-15}$ C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know force for mass

force = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 × $10^{9}$ )

so charges of the beads = 1.173 × $10^{-15}$ C

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2 years ago

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natita [175]

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2 years ago

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Answer:

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Explanation:

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3 years ago