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Triss [41]
1 year ago
6

I need help with this question how to solve it for Brass and Cooper

Physics
1 answer:
Ksenya-84 [330]1 year ago
3 0

Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

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Hoochie [10]

The astronaut will move at 0.4 m/s in the opposite direction to the wrench

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the astronaut-wrench system must be conserved before and after the launch.

Before the launch, the total momentum is zero, since the astronaut is at rest:

p = 0 (1)

After the launch, the total momentum is:

p=mv+MV (2)

where :

m = 2 kg is the mass of the wrench

v = 10 m/s is the velocity of the wrench

M = 50 kg is the mass of the astronaut

V is the recoil velocity of the astronaut

Since momentum is conserved, we can write (1) = (2), and so we can solve for V:

0=mv+MV\\V=-\frac{mv}{M}=-\frac{(2)(10)}{50}=-0.4 m/s

And the negative sign means that the astronaut will move in the opposite direction to the wrench.

Learn more about conservation of momentum:

brainly.com/question/7973509

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8 0
3 years ago
By looking at the relative positions of the elements calcium, Ca, fluorine, F, sulfur, S, and oxygen, O, in the Periodic Table,
SVEN [57.7K]

Answer:

id say the first option.

Explanation:

hope this helps you!

8 0
2 years ago
Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is
Zigmanuir [339]

Answer:

Part a)

I = 1.5 kg m^2

Part b)

I = 0.75 kg m^2

Part c)

I = 1.5 kg m^2

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

I = mL^2 + mL^2 + m(0) + m(0)

I = 2mL^2

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2

I = 2m\frac{L^2}{2}

I = (3 kg)(0.50^2)

I = 0.75 kg m^2

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2

I = 4m\frac{L^2}{2}

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

8 0
3 years ago
Thomas edison. what is the invention behind him?
matrenka [14]
The light bulb... i think ... 
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A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter
lutik1710 [3]

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72  m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion (v_t=\omega\,R=\omega\,(0.25\,m)\,) - with \omega being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}

Then the addition of these two velocities equals:

\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}

7 0
3 years ago
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