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3 years ago

According to Newton’s first law of motion, a moving object that is not acted on by an unbalanced force will..

Physics

2 answers:

damaskus [11]3 years ago

8 0

I believe it’s stay in motion if it’s not acted on by an unbalanced force

exis [7]3 years ago

5 0

Answer:

continue in its motion

Explanation:

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The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

erica [24]

Answer:

0.235 nC

Explanation:

Given:

$E$ = the magnitude of electric field = $8.50\ kN/C =8.50\times 10^{3}\ N/C$

$F$ = the magnitude of electric force on each antenna = $2.00\ \mu N =2.00\times 10^{-6}\ N$

$q$ = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

$\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC$

Hence, the value of q is 0.235 nC.

4 0

3 years ago

What is the lowest possible temperature

HACTEHA [7]

The lowest possible temperature is absolute zero. However scientists have not reached this temperature, rather they have come very close to absolute zero.

6 0

2 years ago

Read 2 more answers

Represent 7468 N with SI units having an appropriate prefix. Express your answer to four significant figures and include the app

Musya8 [376]

Answer:

7.468 kN

Explanation:

Here the force is given in Newton

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

Tera = 10¹²

The number is 7468.0

Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.

1 kilonewton = 1000 Newton

$1\ Newton=\frac{1}{1000}\ kilonewton$

$\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton$

So, 7468 N = 7.468 kN

7 0

3 years ago

Directions: Read the following excerpts carefully, then identify whether they employ informative, persuasive, or argumentative w

Savatey [412]

Hi there!

Informative writing has the intent to inform or educate us on a particular topic or event. It gives us more information and insight onto something.

Persuasive writing has the intent of convincing us to believe in a certain idea or to perform a certain action. For instance, advertisements have a persuasive intent; they are persuading us to buy a product or service.

Argumentative writing is similar to persuasive writing in the sense that they are persuading us to believe a certain idea. However, they are often based on logic and fact rather than opinions.

Let's look at the first excerpt.

<em>This morning at 9 a.m., a school bus collided with a car at the intersection of Osmena and Cabrera streets. The passengers were not injured, but the medical personnel checked each student as well as the driver before they were transported to their school.</em>

This text doesn't try to convince us in believing something. It doesn't argue anything and it only tries to give us more insight onto the event, which is a car accident. No opinions are stated and only events are given.

Therefore, this excerpt uses an informative writing technique.

4 0

2 years ago

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago