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4 years ago

16. Compared to your weight and mass on Earth, if you were on the moon: *

Physics

1 answer:

Natalija [7]4 years ago

5 0

Answer:

The answer is D.

Explanation:

There is no gravity in Space so that means that it will decrease your weight but not your mass.

<h2><u><em>Please give Brainiest</em></u></h2>

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Convert 800 cm to meters.

Rasek [7]

Answer:8 meters

Explanation:100cm is a meter

8 0

2 years ago

Read 2 more answers

The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he

VashaNatasha [74]

Answer:

dV/dt = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

$V = \pi r^{2}h$

put r = h/2 so,

$V = \pi \frac{h^{3}}{4}$

Differentiate both sides with respect to t.

$\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}$

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

$\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3$

dV/dt = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0

3 years ago

A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the end of the

Strike441 [17]

Answer:

c. 2 m/s

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by:

$v=f \lambda$

where

v is the speed of the wave

f is its frequency

$\lambda$ is the wavelength

For the transverse wave in this problem, we have:

$f = 4 Hz$ is the frequency

$\lambda=0.5 m$ is the wavelength

Substituting these numbers into the equation, we find the speed of the wave:

$v=(4 Hz)(0.5 m)=2 m/s$

7 0

3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across

NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:

$A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\$

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>

L= 2.00 m

From Table Copper Resistivity $p$ = 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

$R=\frac{pL}{A}$

=0.0165Ω

The Potential difference across the copper wire is:

V=IR

=2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:

$R=\frac{pL}{A}$

=0.014Ω

The Potential difference across the Silver wire is:

V=IR

=1.75*10^-4

4 0

3 years ago