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3 years ago

The atomic number of a element if found by countingthe number of what in an atom?

1 answer:

4 0

The Atomic number is found by counting the number of Protons found in the atom. Make brainliest if I helped best please :)

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State the name in words and the symbol for the following:

BartSMP [9]

Answer:

<h2>Kilometer (km) and micrometer (um) respectively</h2>

Explanation:

<h3>One thousand meters is equal to one kilometer represented as km. </h3>

and

<h3>One thousandth of a meter mean 1/1000 m which implies one thousands part of a meter which is equal to micro meter and represented as um.</h3>

3 0

3 years ago

What is one way to describe elements

Talja [164]

Answer:

They are the smallest pancise of any matere

4 0

3 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

3 years ago

To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 cm makes 621 vibrations per

PilotLPTM [1.2K]

Answer:

T=9.4 N

Explanation:

We are given that

Mass of wire,m=16.5 g= $\frac{16.5}{1000}$ kg

1 kg=1000g

Length of wire,l=75 cm= $75\times 10^{-2}$ m

1 m=100 cm

Wavelength of transverse wave= $\lambda=3.33 cm=3.33\times 10^{-2}$ m

Frequency= $621 Hz$

Mass per unit length= $m_l=\frac{m}{l}=\frac{16.5}{1000\times 75\times 10^{-2}}=0.022 kg/m$

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

$v=\nu \lambda$

Where $\nu=$ frequency of wave

$\lambda$ =Wavelength of wave

Speed of wave=v

Using the formula

$v=3.33\times 10^{-2}\times 621=20.7m/s$

$v=\sqrt{\frac{T}{m_l}}$

$v^2=\frac{T}{m_l}$

$T=v^2m_l$

Using the formula

$T=(20.7)^2\times 0.022=9.4N$

Hence, the tension,T=9.4 N

7 0

3 years ago

Two golf balls with the same mass were hit at the same time. Golf ball A went a shorter distance than golf ball B. What could be

Firdavs [7]

Answer:

Golf ball A was hit with a weaker force

Explanation:

Golf ball B went a larger distance than golf ball A, so we could think friction affected golf ball A rather than golf ball B. That is why the first two choices are wrong.

The strongest forced applied made golf ball B go further.

7 0

3 years ago

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