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3 years ago

What is the density of iron if 5.0 cm3 has a mass of 39.5g

Physics

1 answer:

weqwewe [10]3 years ago

7 0

Answer : $\rho = 7.9\ g/cm^3$

Explanation :

It is given that

Mass of iron, m = 39.5 g

Volume of iron, $V = 5\ cm^3$

So, density is :

density, $\rho =\dfrac{mass}{volume}$

$\rho =\dfrac{39.5g}{5cm^3}$

$\rho = 7.9\ g/cm^3$

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A car is driven off a cliff at 39 m/s. It lands 141 m from the base. How high

Mumz [18]

Answer: A

Explanation: STEP BY STEP

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2 years ago

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

drek231 [11]

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

Now we have to calculate the heat change of cold water.

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature = $33.4^oC$

Now put all the given value in the above formula, we get:

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$

4 0

3 years ago

A 100-kg astronaut is floating in outer space. If the astronaut throws a 2-kilogram wrench at a speed of 10 meters per second, w

sergey [27]

Since Astronaut and wrench system is isolated in the space and there is no external force on it

So here momentum of the system will remain conserved

so here we can say

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

initially both are at rest

so here plug in all values

$0 = 100 v_{1f} + 2\times 10$

$v_{1f} = -0.20 m/s$

so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s

3 0

3 years ago

The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it

soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

$U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})$

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, $v_a$ as 0, 9.81 for g then

$58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s$

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3 years ago

The head of a rattlesnake can accelerate at 49 m/s2 in striking a victim. If a car could do as well, how long would it take to r

Anettt [7]

<h2>Time taken is 0.459 seconds</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 81 km/hr = 22.5 m/s

Time, t = ?

Acceleration, a = 49 m/s²

Substituting

v = u + at

22.5 = 0 + 49 x t

t = 0.459 seconds

Time taken is 0.459 seconds

3 0

3 years ago