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3 years ago

An IGCSE student thinks it may be possible to identify different rocks (A, B and C) by measuring their

Physics

1 answer:

kherson [118]3 years ago

8 0

Answer:

See the answer below

Explanation:

a. The volume in the first measuring cylinder reads 70 $cm^3$ while that of the second reads 95 $cm^3$ . Hence;

V1 = 70 $cm^3$

V2 = 95 $cm^3$

b. An object will always displace its own volume in a liquid. Hence:

Volume V of the rock sample = V2 - V1

= 95 - 70 = 25 $cm^3$

c. Mass of A = 102 g

Volume of A = 25 $cm^3$

Density = mass/volume

Hence, density of A = 102/25 = 4.08 g/ $cm^3$

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Pulleys accomplish 2 separate operations throughout the computer controlled additional benefit technologies listed elsewhere here.

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2 years ago

Define neuron and nerve cell

san4es73 [151]

Answer: Hi!

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2 years ago

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How many number of musicians constitute a big band?

nikitadnepr [17]

Big band is music group (a group of people who perform instrumental and/or vocal music ) playing jazz or jazz-influenced popular music and which was popular during the Swing Era from the mid-1930s until the late 1940s. These big bands contained saxophones, trumpets, trombone and other instruments and typically consisted of approximately 12 to 25 musicians.

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3 years ago

In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the

Vedmedyk [2.9K]

Answer:

2.2 s

Explanation:

Hi!

Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

$x(t) = 23.1 m - \frac{1}{2}(9.8 m/s^2) t^2$

We are looking for a time t for which x(t) = 0

$0 = 23.1 m - (4.9 m/s^2) t^2$

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2 years ago

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m

Nastasia [14]

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by: $W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$ )

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ( $6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$ )

$m_{1} , m_{2}$ = masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

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3 years ago