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3 years ago

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A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____

_.

1 answer:

Rama09 [41]3 years ago

5 0

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$ , f = focal length of the mirror

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}$

v = -8.57 cm

The magnification of a mirror is given by,

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-8.57)}{-20}$

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

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Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic

marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = $1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

Q = 0.063 m³/s

flow rate in liters per minute

1 gallon = 3.78541 L

Q = $1000\times \dfrac{3.78541\ m^3}{1\ gallon}$

Q = 3785.41 m³/min

flow rate in cubic feet per second

1 gallon = 0.133681 ft³

1 min = 60 s

Q = $1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

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3 years ago

A soccer player kicks a soccer ball with a force of 1.8 N. If the mass of the ball is .43 kg. How fast will the ball accelerate?

Vesnalui [34]

Answer:

The acceleration of the ball is 4.18 [m/s^2]

Explanation:

By Newton's second law we can find the acceleration of the ball

$F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]$

Now we have:

$a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]$

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3 years ago

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One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

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3 years ago

Tectonic plates are large segments of the earth's crust that move slowly. suppose one such plate has an average speed of 4.8 cm

user100 [1]

<span>one year is 365, 1 day is 24 hours, 1 hour is 60 minutes, 60 minutes is 60 seconds, thus (365 * 24 * 60 * 60) = 31,536,000 one year is equal to 31,536,000 seconds. the plate has a speed of 4.8 cm every 31,536,000 seconds. lets find out how far it goes in 40 seconds. (4.8/31,536,000)*40 = 0.00000608828 The plate moves 0.00000608828 cm every 40 seconds</span>

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3 years ago

A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl

Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$

$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$

Part C

The error in % is given by

$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$

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3 years ago

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