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2 years ago

1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.

Physics

1 answer:

Mashutka [201]2 years ago

4 0

The position of the object is = -68cm

The magnification of the mirror= 0.3

<h3>Calculation of object distance</h3>

The image distance = 20.5cm

The focal length= R/2 = 31.5/2= 15.75

The object distance= ?

Using the lens formula,1/f = 1/v-1/u

1/u = 1/v- 1/f

1/u = 1/20.5 - 1/15.75

1/u = 0.0489- 0.0635

1/u = -0.0146

u = -68cm

The magnification of the mirror is image size/object size

= 20.5cm/-68cm

= 0.3

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A pharmacist works with a 1.75 m solution of sodium bromide (nabr) and water. the volume of the solution is 84.0 milliliters. if

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The volume of the diluted (i.e the new) solution given the data is 147 mL

<h3>Data obtained from the question </h3>

Molarity of stock solution (M₁) = 1.85 M
Volume of stock solution (V₁) = 84 mL
Molarity of diluted solution (M₂) = 1 M
Volume of diluted solution (V₂) = ?

<h3>How to determine the volume of diluted solution </h3>

M₁V₁ = M₂V₂

1.75 × 84 = 1 × V₂

V₂ = 147 mL

Thus, the volume of the new solution is 147 mL

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2 years ago

An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof

Pavel [41]

Answer:

$T = 69^o C$

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

$Q_{ice} = Q_{coffee}$

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

$m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2$

here we have

$m_1 = 15 gram$

L = 333 kJ/kg = 333 J/g[/tex]

$c_1 = c_2 = 4186 J/kg C = 4.186 J/g C$

$\Delta T_1 = T - 0$

$\Delta T_2 = 85 - T$

now we have

$15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)$

$4995 + 62.79T = 49813.4 - 586.04T$

$648.83 T = 44818.4$

$T = 69^o C$

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3 years ago

Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitud

navik [9.2K]

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, $q_1=-6.29\times 10^{-6}\ C$

Second charged particle, $q_2=5.23\times 10^{-6}\ C$

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}$

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

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3 years ago

A golfer hits a golf ball at an angle of 25 degrees to the ground at a speed of 76 m/s. If the gold ball covers a horizontal dis

Lady bird [3.3K]

The ball's horizontal position x and vertical position y at time t are given by

x = (76 m/s) cos(25º) t

y = (76 m/s) sin(25º) t - 1/2 g t²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball's initial vertical velocity is (76 m/s) sin(25º) ≈ 32.12 m/s.

Its initial horizontal velocity is (76 m/s) cos(25º) ≈ 68.88 m/s.

The ball stays in the air for as long as y > 0. Solve y = 0 for t :

(76 m/s) sin(25º) t - 1/2 g t² = 0

t ((76 m/s) sin(25º) - 1/2 g t ) = 0

t = 0 or (76 m/s) sin(25º) - 1/2 g t = 0

Ignore the first solution.

(76 m/s) sin(25º) - 1/2 g t = 0

(76 m/s) sin(25º) = (4.90 m/s²) t

t = (76 m/s) sin(25º) / (4.90 m/s²)

t ≈ 6.55 s

Recall that

v² - u² = 2 a ∆y

where u and v denote initial and final velocities, a is acceleration, and ∆y is displacement. At maximum height, the ball has zero vertical velocity, and taking the ball's starting position on the ground to be the origin, ∆y refers to the maximum height. So we have

0² - ((76 m/s) sin(25º))² = 2 (-g) ∆y

∆y = ((76 m/s) sin(25º))² / (2g)

∆y ≈ 52.6 m

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3 years ago

Burning fuel _______________ carbon dioxide into the atmosphere. *

Igoryamba

Answer:

Releases

Explanation:

Burning fuel releases carbon dioxide into the atmosphere. This is because when the fuel burns, the toxic air escapes it and flows into the atmosphere and eventually dissipates into what we see as pollution.

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3 years ago