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3 years ago

Find your mass if a scale on earth reads 650 N when you stand on it.

1 answer:

7 0

Weight = (mass) x (acceleration of gravity) .

On Earth, acceleration of gravity is 9.8 m/s² (rounded) .

650 N = (mass) x (9.8 m/s²)

Divide each side by (9.8 m/s²): 650 N / 9.8 m/s² = mass

Mass = 66.3 kilograms (rounded)

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The speed of sound through diamond is about 12,000 m/s. The speed of sound through wood is about 3,300 m/s. Which statement expl

Damm [24]

Answer: They have different rigidities.

Explanation:

7 0

2 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

3 years ago

A research submarine has a 20-cm-diameter window that is 9.0 cm thick. The manufacturer says the window can withstand forces up

Ratling [72]

Answer:

The maximum safe depth in salt water is 3758.2 m.

Explanation:

Given that,

Diameter = 20 cm

Radius = 10 cm

Thickness = 9.0 cm

Force $F = 1.2\times10^{6}\ N$

Inside pressure = 1.0 atm

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

Put the value into the formula

$A=\pi\times(10\times10^{-2})^2$

$A=0.0314\ m^2$

We need to calculate the pressure

Using formula of pressure

$P=\dfrac{F}{A}$

Put the value into the formula

$P=\dfrac{1.2\times10^{6}}{0.0314}$

$P=38216560.50\ Pa$

$P=3.8\times10^{7}\ Pa$

We need to calculate the maximum depth

Using equation of pressure

$P=P_{atm}+\rho gh$

$h=\dfrac{P-P_{atm}}{\rho g}$

Put the value into the formula

$h=\dfrac{3.8\times10^{7}-101325}{1029\times9.8}$

$h=3758.2\ m$

Hence, The maximum safe depth in salt water is 3758.2 m.

4 0

3 years ago

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T on Earth. Calculate the

melamori03 [73]

Answer:

T = T

Explanation:

Time period of a simple pendulum is not affected by the mass of the bob. As we know, $T = 2\pi \sqrt{L/g}$ . There is no factor of mass affecting when we derived the equation. The basic reason behind the time period is not affected is because of mass dependence on angular acceleration. As the mass increases the acceleration increase and the Time Period remains constant.

8 0

3 years ago

The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad

11111nata11111 [884]

Answer:

(a) $Q = 7.28\times 10^{14}$

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

$\int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\Q_enc = 7.28\times 10^{14}$

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

5 0

3 years ago