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Mila [183]
3 years ago
6

A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th

e mirror concave or convex? What is its radius of curvature?
Physics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

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A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The
oksian1 [2.3K]

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

\mu mg = \frac{mv^2}{R}

here we have

\mu g = \frac{v^2}{R}

now we know that

v = \sqrt{\mu Rg}

here we have

\mu = 0.600

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

v = \sqrt{(0.600)(35)(9.81)}

v = 14.35 m/s

4 0
3 years ago
An object with a momentum of 1500 kg-m/s directed east is acted upon by an impulse of 100.0 kg-m/s in the same direction. What i
fiasKO [112]

Answer:

The final momentum of the body = 1600 kgm/s

Thus the impulse that acted on the body. = 500 N.s

Explanation:

Momentum: This can be defined as the product of mass and velocity. The S.I unit of momentum is kgm/s. Momentum is a vector quantity.

Mathematically, momentum can be expressed as

M =mv.

Where M = momentum of the object, m = mass of the object, v = velocity of the object.

Impulse acting on the object = Final momentum of the object - initial momentum of the object

I = M₂ - M₁

M₂ = I + M₁......................... Equation 1

Where I = impulse, M₁ and M₂ = Final and initial momentum.

Note:

(i) The momentum and impulse act in the same direction

(ii) impulse is also a vector quantity.

Given: M₁= 1500 kgm/s, I = 100 kgm/s.

Substituting these values into equation 1

M₂ = 1500 + 100

M₂ = 1600 kgm/s.

Thus the final momentum of the body = 1600 kgm/s

14.

I = m(v-u)............................................... Equation 2.

Where I = impulse on the object, m = mass of the object, v = final velocity, u = initial velocity.

Given: m = 100 kg, v = 15 m/s, u = 10 m/s.

I = 100(15-10)

I = 100(5)

I = 500 N.s or 500 kgm/s.

Thus the impulse that acted on the body. = 500 N.s

7 0
4 years ago
A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.
REY [17]

Answer: 85.46\ kJ

Explanation:

Given

Volume of air V=105\ m^3

Temperature of air T=305\ K

Increase in temperature \Delta T=0.7^{\circ}C

Specific heat for diatomic gas is C_p=\dfrac{7R}{2}

Energy required to increase the temperature is

\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]

Insert the values

\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ

6 0
3 years ago
What will happen to a periodic wave acted upon an external damping force?
OverLord2011 [107]
The wave will decrease in its frequency due to a disturbing force acting upon it.
4 0
4 years ago
Read 2 more answers
Which remains the same as the distance of an object from Earth changes?
Gemiola [76]

Answer:

MASS

Explanation:

weight has to do with the gravitational pull, as distance from earth decreases, so does the gravitational pull. meaning, the size of the force, the pull, and the weight would all change. mass stays the same (in this sense. if you gain weight on earth you will gain mass as well, but if you leave earth your weight will lessen and your mass will stay the same.)

hope this helped.

5 0
4 years ago
Read 2 more answers
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