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3 years ago

12

Please help me with my quiz

1 answer:

Oduvanchick [21]3 years ago

8 0

Answer:

matter: A

data: E

variable: C

Controlled: D

Physical: B

Explanation:

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Answer the question fast please

AnnZ [28]

Answer:

1

Explanation:

$d = m \div v \\ m = 0.8 \: v = 80\% \: also \: 0.8 \\ d = 0.8 \div 0.8 \\ d = 1$

6 0

3 years ago

Draw sodium formate by placing atoms on the grid and connecting them with bonds. Include all lone-pair electrons.

nikklg [1K]

Answer:

Sodium formate is the sodium salt of formic acid which is given as HCOONa.

Explanation:

The basic structure of Sodium formate consists of following bonds:

The main Ionic bond between the $HCOO^-$ radical and $Na^+$ .
The sigma covalent bonds between atom of H, atom of C and both atoms of O.
The pi bond between atom of C and one atom of O.

The structure along with lone pairs is given as attached

7 0

3 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

4 years ago

The auroras occur in the

kotykmax [81]

Answer:

Ionosphere

Explanation:

The thermosphere reaches 600 kilometres just above mesosphere and begins immediately above the mesosphere. This layer is where the aurora and satellites appear.

The ionosphere is the comprehensive career of the mesosphere because most of the thermosphere, located 80–400 kilometres just above ground atmosphere.

Auroras — magnificent flowing streaks of light seen in the night sky – appear in this location.

8 0

3 years ago

A flute player hears four beats per second when she compares her note to a 523 Hz tuning fork (the note C). She can match the fr

xxTIMURxx [149]

Answer:

527Hz

Explanation:

The beat frequency of any two waves is:

Fbeat =|f1 – f2|

i.e

for frequency tied to length increase (wavelength increases as length increases, therefore frequency decrease:: Fbeat = f1+f2

for frequency tied to length decrease wavelength increases as length increases, therefore frequency also increases : Fbeat = f1-f2

Note that a wavelength increase means a decrease of frequency because v = fλ

Therefore from the question:

Fbeat =4 beats/s

F2 -523Hz

Fbeat = f1-f2

F1=Fbeat+F2

=523+4

=527Hz

6 0

3 years ago