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2 years ago

14

...please help me :,)

1 answer:

natka813 [3]2 years ago

4 0

it doesnt show a picture or anything for me soorryy

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A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh

lianna [129]

Answer: 3.21 N

$Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}$

$\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg$

For weight, we will multiply by $g=9.8 m/s^{-2}$

$weight= 0.328\times9.8=3.21\hspace{1mm}N$

Hence, the rock would weigh 3.21 N.

3 0

2 years ago

The block in the drawing has dimensions L0×2L0×3L0, where L0 =0.5 m. The block has a thermal conductivity of 250 J/(s·m·C˚). In

stira [4]

Answer:

I am sorry but what drawing

3 0

2 years ago

During the annual shuffleboard competition, Renee gives her puck an initial speed of 8.12 m/s. Once leaving her stick, the puck

shutvik [7]

<h2>Answer:2.65 seconds</h2>

Explanation:

Let $a$ be the acceleration.

Let $u$ be the initial velocity.

Let $v$ be the final velocity.

Let $t$ be the time taken.

As we know from the equations of motion,

$v=u+at$

Given, $v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}$

$0=8.12-3.06t$

$t=2.65sec$

6 0

3 years ago

Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%

djyliett [7]

Answer:

$6.8370869499\times 10^{20}\ N$

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

e = Charge of electron = $1.6\times 10^{-19}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

$q=imbalance\times n\times e$

Total number of protons and electrons in each sphere

$n=\dfrac{mN_AZe}{M}$

$q=imbalance\times \dfrac{mN_AZe}{M}$

$q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C$

$q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C$

Electrical force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N$

The electrostatic force of attraction between them is $6.8370869499\times 10^{20}\ N$

4 0

3 years ago

Help, It has multiple answers for this question.

evablogger [386]

The answer is ultra violet radiation. From the air

4 0

3 years ago