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Bess [88]
3 years ago
7

How long is the image of a metrestick formed by a plane mirror?

Physics
1 answer:
Zinaida [17]3 years ago
7 0

A plane mirror doesn't change the apparent size or shape of things.

-- If the meter stick is held parallel to the mirror, or is lying on the mirror,
then its image in the mirror is 1 meter long.

-- If the meter stick is held perpendicular to the mirror, or is standing on
the mirror, or pointing directly at it, then its image in the mirror has no length.

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If 10. joules of work must be done to move 2.0 coulombs of charge from point A to point B in an electric field, the potential di
masya89 [10]
The equation for work (W) done by an electric field is:

W = qΔV

where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:

10 = 2ΔV
ΔV = 5
8 0
3 years ago
suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?
guajiro [1.7K]
<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>
6 0
3 years ago
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Which of the following terms is correct when talking about energy?
vlabodo [156]

Answer:

C. Converting Energy

Explanation:

Hope this helped, Have a Wonderful Day!!

5 0
2 years ago
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What is a temperature where the vapor pressure of a liquid equals the atmospheric pressure.
AnnZ [28]
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5 0
3 years ago
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

8 0
3 years ago
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