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2 years ago

5

A sphere of metal has a mass of 1431 g and a diameter of 5.80 cm. What is the density of metal in g/cm3

1 answer:

Dmitriy789 [7]2 years ago

4 0

Answer:

see below

Explanation:

You will need t find the volume of the sphere

4/3 pi r^3 divide into the mass

1431 / (4/3 pi (5.8)^3) = 14 gm /cm^3

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A crate is dragged 4.0 m along a rough floor with a constant velocity by a worker applying a force of 400 N to a rope at an angl

Harman [31]

For the crate moving along the direction of Force with constant velocity the summation of the force alng x-axis should be zero. and too for the y component
In view of this
Fx=Fcos30
=364.4N

8 0

3 years ago

A 100 Ω resistor is connected in series with a 47 µF capacitor and a source whose maximum voltage is 5 V, operating at 100.0 Hz.

Pani-rosa [81]

Answer:

$X_c=-33.86275385\Omega$

$|Z|=105.5778675\Omega$

$I=0.04735841062A$

$\phi=20.78612878\°$

Explanation:

The electrical reactance is defined as:

$X_c=-\frac{1}{2\pi fC}$

Where:

$f=Frequency\\C=Capacitance$

So, replacing the data provided by the problem:

$X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega$

Now, the impedance can be calculated as:

$Z=R+jX_c$

Where:

$R=Resistance\\X_c= Capacitive\hspace{3}reactance$

Replacing the data:

$Z=100-j33.86275385$

In order to find the magnitude of the impedance we can use the next equation:

$|Z|=\sqrt{(R^2)+(X_c^2)}=\sqrt{(100)^2+(-33.86275385)^2} =105.5778675\Omega$

We can use Ohm's law to find the current:

$V=I*Z\\I=\frac{V}{Z}$

Therefore the current is:

$I=\frac{5}{100-j33.86275385}=0.04485638113+0.01518960593j$

And its magnitude is:

$|I|=\sqrt{(0.04485638113)^2+(0.01518960593)^2} =0.04735841062\Omega$

Finally the phase angle of the current is given by:

$\phi=arctan(\frac{0.01518960593}{0.04485638113})=20.78612878\°$

5 0

3 years ago

(a) The stick is supported by a sharp point at the middle. On the left side, a weight of 100 g is suspended at 40 cm from the mi

Jet001 [13]

Answer:

20cm

Explanation:

Hello!

remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,

To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation

(100g)(40cm)=x(200g)

$X=\frac{(100)(40)}{(200)} =20cm$

6 0

3 years ago

A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a

9966 [12]

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination $\theta$ = 15°

$\sin \theta = \frac{AB}{AC}$

$\sin 15 = \frac{5900}{x}$

$x = \frac{5900}{0.2588}$

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

7 0

3 years ago

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

8 0

3 years ago