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2 years ago

Gold has a density of 19.3 g/cm3. I fyou has a 2.00 cm3 block of gold, what would be its mass?

Physics

1 answer:

dalvyx [7]2 years ago

4 0

Density equals mass × volume

d=m/V

then m=d×V=19.3×2=38.6g

m=38.6g

hope this helps you

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You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment

Margarita [4]

Answer:

Explanation:

We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .

14 m due west = - 14 i

22.0 m upward in the elevator = 22 k

12 m north = 12 j

6.00 m east = 6 i

Total displacement = - 14 i + 22 k + 12 j + 6 i

D = - 8 i + 12 j + 22 k

magnitude = √ ( 8² + 12² + 22² )

= √ ( 64 + 144 + 484 )

= √ 692

= 26.3 m

Net displacement from starting point = 26.3 m .

5 0

2 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

A pelican flying along a horizontal path drops

Ludmilka [50]

Answer:

The initial speed of the pelican is 8.81 m/s.

Explanation:

Given;

height of the pelican, h = 5.0 m

horizontal distance, X = 8.9 m

The time of flight is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*5}{9.81} } \\\\t = 1.01 \ s$

The initial horizontal speed of the pelican is given by;

X = vₓt

vₓ = X / t

vₓ = 8.9 / 1.01

vₓ = 8.81 m/s

Therefore, the initial speed of the pelican is 8.81 m/s.

8 0

3 years ago

* CRIMINOLOGY*

Setler [38]

I think the answer is B.

Hope this helps.

5 0

3 years ago

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What are the 4 energy transformations of a dash toy

Hunter-Best [27]

Answer:

Thermal/Heat energy, kinetic energy, light energy, & Electromagnetic energy

3 0

2 years ago