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3 years ago

A lever in which the load is between the fulcrum and the applied force is a ________.

1 answer:

4 0

Hello there.

<span>A lever in which the load is between the fulcrum and the applied force is a ________.

Answer: It is a second class lever.

Hope This Helps You!
Good Luck Studying ^-^</span>

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A parallel-plate capacitor is connected to a battery until it is fully charged. Then, the capacitor is disconnected from the bat

Helga [31]

Answer:

The potential between the plates will decrease.

Explanation:

An insulator is usually placed between the parallel plates and is also called a dielectric because it makes the amount of charge a capacitor can accommodate to increase at a particular potential difference.

Furthermore, the dielectric effect will make the electric field of the charged capacitor which is not connected to a source of supply to decrease.

Now, when the battery is removed, the charge Q remains constant and Capacity C will increase.

Formula for the potential difference is here;

V = Q/C

Since the numerator Q is constant and the denominator C increases, it means the potential difference V will decrease

4 0

3 years ago

The force required to stretch a Hooke’s-law

Setler [38]

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K

Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J

5 0

3 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. find the minimum en

shusha [124]

The energy carried by the incident light is
$E=hf$
where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
$E=hf=(6.63 \cdot 10^{-34}Js)(5.64 \cdot 10^{14}s^{-1})=3.74 \cdot 10^{-19}J$

4 0

3 years ago

Humpback whales are known to produce a collection of elaborate and repeating sounds with frequencies starting at 20 Hz. The soun

jasenka [17]

Answer:

70 m.

Explanation:

Given,

Frequency, f = 20 HZ

speed of sound, v = 1400 m/s

wavelength of the waves = ?

we know,

v = f λ

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

$\lambda= \dfrac{1400}{20}$

$\lambda=70\ m$

Hence, the wavelength of the wave is equal to 70 m.

8 0

3 years ago

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person

sertanlavr [38]

Answer:

v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

f ’= f (v + v₀) / (v- $v_{s}$ )

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

v₀ = v_{s} = v ’

we substitute

f ’= f (v + v’) / (v - v ’)

f ’/ f (v-v’) = v + v ’

v (f ’/ f -1) = v’ (1 + f ’/ f)

v ’= (f’ / f-1) / (1 + f ’/ f) v

v ’= (f’-f) / (f + f’) v

let's calculate

v ’= (3400 -3000) / (3000 +3400) 343

v ’= 400/6400 343

v ’= 21.44 m / s

3 0

2 years ago