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2 years ago

Consider a concave spherical mirr or that has focal length f = +19.5 cm.

1 answer:

7 0

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

f is the focal length of the mirror
v is the object distance
u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

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False

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2 years ago

A elephant kicks a 5.0\,\text {kg}5.0kg5, point, 0, start text, k, g, end text stone with 150\,\text J150J150, start text, J, en

S_A_V [24]

The speed of the stone is 7.7 m/s

Explanation:

The kinetic energy of a body is the energy possessed by the body due to its motion. Mathematically,

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For the stone in this problem, we have:

K = 150 J is its kinetic energy

m = 5.0 kg is its mass

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$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(150)}{5.0}}=7.7 m/s$

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3 years ago

Pre-questioning identifies a purpose for reading.

nadya68 [22]

Answer:

True

Explanation:

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3 years ago

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A helium-filled balloon is launched when the temperature at ground level is 27.8°c and the barometer reads 752 mmhg. if the ball

ololo11 [35]

The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.

Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³

Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K

At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K

If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³

Answer: 762.2 m³

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3 years ago

250 mL 0.1 M HCl solution is mixed with 250 mL

pishuonlain [190]

Answer:

The concentration of OH⁻ in the mixture is 0.05 M

Explanation:

The reaction of neutralization between HCl and NaOH is the following:

H⁺(aq) + OH⁻(aq) ⇄ H₂O(l)

The number of moles of HCl is:

$n_{HCl} = C*V = 0.1 mol/L*0.250L = 0.025 moles$