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2 years ago

14

calculate the load placed 10m from the fulcrum that can be balanced by an effort of 5 N applied at a distance of 4 m from the fu

lcrum in a lever

1 answer:

Elena L [17]2 years ago

8 0

Answer:

A = 2 m from fulcrum

Explanation:

Product of anti clockwise = Product of clockwise moment

5 × 4 = 10 × A

20 = 10 x A

A = 20 / 10

A = 2 m from fulcrum

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A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

$V = \sqrt{\frac{T}{\rho}}$

Where,

T = Tension

$\rho =$ Linear density

Our data are given by

Tension , T = 70 N

Linear density , $\rho = 0.7 kg/m$

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

$V = \sqrt{\frac{T}{\rho}}$

$V = \sqrt{\frac{70}{0.7}$

$V = 10m/s$

Speed can also be expressed as

$V = \lambda f$

Re-arrange to find \lambda

$\lambda = \frac{V}{f}$

Where,

f = Frequency,

Which is also described in function of the Period as,

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

$f = \frac{1}{0.35}$

$f = 2.86 Hz$

Therefore replacing to find $\lambda$

$\lambda = \frac{10}{2.86}$

$\lambda = 3.49m$

Therefore the wavelength of the waves created in the string is 3.49m

3 0

3 years ago

What is the density of a piece of wood with a mass of 25 kg<br> and a volume of 0.0385 m³?

Orlov [11]

Answer:

649kg/m^3

Explanation:

Let p be the density of this particular object.

Formula for density:

$p = \frac{mass \: (in \: kg)}{volume \: (in \: {m}^{3}) }$

We can substitute the givenmass and volume to find density of the object.

$p = \frac{25kg}{0.0385 {m}^{3} } \\ = 649kg \: per \: {m}^{3}$

Therefore the density of this object is 649kg/m^3.

7 0

1 year ago

all objects near the earths surface-regardless of size and weight have the same force of gravity acting on them. is it true or f

pickupchik [31]

False: the force of gravity acting on different objects is different and depends on their mass

Explanation:

The answer is false.

The force of gravity acting on an object (also known as weight) near the Earth's surface is given by:

$F=mg$

where:

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

We see from the formula that the force of gravity acting on an object depends on the mass: the larger the mass of the object, the stronger the gravitational force acting on it, and the smaller the mass, the weaker the force of gravity.

The factor that does not change is the acceleration of gravity, which is constant ( $9.8 m/s^2$ ) if we are near the Earth's surface, and implies that all the objects in free fall accelerate at the same rate towards the ground, regardless of their size and weight.

Learn more about forces and weight here:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

5 0

3 years ago

The voltage ???? in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance ???? is slowly inc

zimovet [89]

Answer:

The change in current at $R =456 \Omega$ is $\frac{dI}{dt} = 7.032 * 10^{-5} A/s$

Explanation:

From the question we are told that

The resistance is $R = 465 \Omega$

The current is $I = 0.09A$

The change in voltage with respect to time is $\frac{dV}{dt} = - 0.03 V/s$

The change in resistance with time is $\frac{dR}{dt} = 0.03 \Omega /s$

According to ohm's law

$V = IR$

differentiating with respect to time using chain rule

$\frac{dV}{dt} = I \frac{dR}{dt} + R * \frac{dI}{dt}$

substituting value at R = 456

$-0.0327 = 0.09 * 0.03 + 456* \frac{dI}{dt}$

$\frac{dI}{dt} = 7.032 * 10^{-5} A/s$

6 0

2 years ago

Juan's mother drives 12.5 miles southwest to her favorite shopping mall. What is the velocity of her

SVETLANKA909090 [29]

Answer:

Calculate using the formula

Explanation:

velocity= displacement (m)/time(s)

1 mile =1.6km

1km=1000m

7 0

2 years ago