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3 years ago

All scientists try to base their conclusions on

Physics

2 answers:

Crank3 years ago

5 0

They base their conclusions off Data that they recorded when conducting experiments.

Dima020 [189]3 years ago

3 0

Mathematics and science

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Uses of convace, convex mirror

Leya [2.2K]

Answer:

The concave mirror is a converging mirror so that it is used for many purposes, It is used as a torch to reflect the light, It is used in the aircraft landing at the airports to guide the aeroplanes, It is used in shaving to get an enlarged and erect image of the face.

Explanation:

3 0

3 years ago

A motorcycle has a mass of 2.50×10^2. A constant force is exerted on it for 60.0s. The motorcycles initial velocity is 6.00 m/s

Margarita [4]

The change in momentum is 5500 kg m/s

Explanation:

The change in momentum of an object is given by

$\Delta p = m(v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have:

$m=2.50\cdot 10^2 kg$ (mass of the motorcycle)

$v=28.0 m/s$ (final velocity)

$u=6.0 m/s$ (initial velocity)

Therefore, the change in momentum is

$\Delta p=(2.50\cdot 10^2)(28.0-6.0)=5500 kg m/s$

Learn more about change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

8 0

3 years ago

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

3 years ago

A truck is traveling at 2.0 m/s. It slows to a stop at a constant rate over 3.00 s. How far does the car travel during those 3.0

earnstyle [38]

Answer:

During those 3.00 seconds before stopping, the car travels a distance of 6 m.

Explanation:

The simple rule of three is a tool that is used to quickly solve problems, where three pieces of information must be known, and one of them operates as an unknown to be known.

Two magnitudes are directly proportional if one magnitude increases the other also does it, and if the magnitude decreases the other in the same way.

Being a, b and c known data and x the unknown, the value that we want to know, the rule of three when the magnitudes are directly proportional is applied as follows:

a ⇒ b

c ⇒ x

So: $x=\frac{c*b}{a}$

In this case, knowing that a truck travels at 2 m/s, the rule of three applies as follows: if in 1 second the truck travels 2 m, in 3 seconds how much distance does it travel?

$distance=\frac{3 s*2 m}{1 s}$

distance= 6 m

<u><em> During those 3.00 seconds before stopping, the car travels a distance of 6 m.</em></u>

8 0

3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

4 years ago

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