All categories

3 years ago

An instrument used to detect a static electric charge is called an

2 answers:

7 0

It is B. false that an instrument used to detect a static electric charge is called an ammeter. It is actually called an electroscope. Ammeter measures current.

Andru [333]3 years ago

6 0

b. false

An ammeter is used to measure current, not static electric charge.

You might be interested in

Two students suggest different functions for the battery plays a flashlight.

irga5000 [103]

Answer:

Student b

Explanation:

Because, Battery pumps the charges around the circuit. Here, charges are the free electrons. And this charges are already present in the conductor. When we connect this conductor with the battery - It set up an electric field and The diffrence between the potential between the ends of the wire pushes this free charges to move in perticular direction. (This is called electric energy)

This electric energy is converted by chemical reactions in the battery. When the battery is dead - It means that the chemical energy producing chemicals has been consumed. thereafter there will be no more chemical reactions takes place to produce electric energy. Hence we say that the battery is dead.

3 0

3 years ago

Read 2 more answers

Laura and Elana are discussing how to solve the following problem: "A canary sits 10 m from the end of a 30-m-long clothesline,

horrorfan [7]

Answer:

$f=2.236\ Hz$

Explanation:

Given:

Length of a rope, $l=30\ m$

Position of Canary on the rope from one end, $l_c=10\ m$

Position of Grackle on the rope from another end, $l_g=5\ m$

Tension in the rope, $F_T=200\ N$

linear mass distribution on the rope, $\mu=0.1\ kg.m^{-1}$

We have for the speed of wave on the string:

$v^2=\frac{F_T}{\mu}$

$v^2=\frac{200}{\0.1}$

$v=44.7\ m.s^{-1}$

For canary to be undisturbed we need a node at this location.

Also, at the end close to Canary there must be a node to avoid any change in pattern of vibration.

So,

the distance between Canary and the closer end must be equal to half the wavelength.

$\frac{\lambda}{2} =10\ m$

$\Rightarrow \lambda=20\ m$

∴Wavelength of wave to be produced = 20 m. This will give us nodes at the multiples of 10 and anti-nodes at the multiples of 5.

Now, frequency:

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

$f=\frac{44.7}{20}$

$f=2.236\ Hz$

3 0

3 years ago

What is good deductive reasoning called

andrezito [222]

to make sure the argument is true and correct backed up by logic.

6 0

3 years ago

what is the magnitude of the gravitational force acting on the earth due to the sun? express your answer in newtons.

Alborosie

The gravitational force the sun experiences from the earth is 3.48×10²²N, which is exactly the same as the force the sun experiences from the earth.

Gravity is a force that develops as a result of the attraction between mass-containing objects. The mass of the object has a direct relationship to the strength of this attraction. r equals the separation of two objects.

F = G (M₁M₂)/r²

Where, F the gravitational force

G=6.67×10⁻¹¹Nm²kg⁻² gravitational constant

M₁=5.98×10²⁴kg mass of earth

M₂= 1.99×10³⁰ kg the mass of the sun

r =15×10¹⁰ m is the distance between sun and earth

Putting all the values in above equation,

F = 6.67×10⁻¹¹Nm²kg⁻²(5.98×10²⁴kg 1.99×10³⁰ kg)/15×10¹⁰ m

On solving the above equation we get,

F = 3.48×10²²N

To know more about gravitational force

brainly.com/question/12830265

#SPJ4

5 0

1 year ago

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

3 years ago