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2 years ago

In the following free body diagram, what is the net force on the object?

Physics

1 answer:

frozen [14]2 years ago

3 0

Answer:

Explanation:

Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.

ΣF = 7 + 18 + (-20) = 5N to the right

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In the 1920s what did Edmund hubble notice about the galaxies

zlopas [31]

Hubble noticed that the galaxies were moving away from us, which meant the universe was expanding.

This is why constellations change over time. In some years, the Big Dipper won't actually look like a dipper anymore.

5 0

2 years ago

Object A has a mass of 100 grams. Object B has a mass of 150 grams. They are both traveling at the same velocity. What can you c

Scilla [17]

-- Momentum is (mass) x (speed).
Object B has 1.5 times as much momentum as Object A has.

-- Kinetic energy is (1/2) x (mass) x (speed) .
Object B has 1.5 times as much kinetic energy as Object A has.

-- If they would both stop long enough to get on the scale,
Object B would weigh 1.5 times as much as Object A does.

8 0

2 years ago

n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

2 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

2 years ago

In some circumstances, it is useful to look at the linear velocity of a point on the blade. The linear velocity of a point in un

mihalych1998 [28]

Answer:

$v=wr$

Explanation:

<u>Tangent and Angular Velocities</u>

In the uniform circular motion, an object describes the same angles in the same times. If $\theta$ is the angle formed by the trajectory of the object in a time t, then its angular velocity is

$\displaystyle w=\frac{\theta}{t}$

if $\theta$ is expressed in radians and t in seconds the units of w is rad/s. If the circular motion is uniform, the object forms an angle $2\theta$ in 2t, or $3\theta$ in 3t, etc. Thus the angular velocity is constant.

The magnitude of the tangent or linear velocity is computed as the ratio between the arc length and the time taken to travel that distance:

$\displaystyle v=\frac{\theta r}{t}$

Replacing the formula for w, we have

$\boxed{ v=wr}$

4 0

3 years ago