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Vesnalui [34]
2 years ago
14

Near the surface of the moon, the distance that an object falls is a function of time. It is given by d(t) = 2.6667t2, where t i

s in seconds and d(t) is in feet. If an object is dropped from a certain height, find the average speed of the object from t = 1 to t = 4. (Round your answer to the nearest whole number.)
Physics
1 answer:
Eddi Din [679]2 years ago
6 0

Answer:

averaage speed  is  v = 13 feet / s

Explanation:

 The average speed is the ratio of the distance traveled in a given time interval, let's calculate the distance that the body travels in the two instants of time that give us

   

 t = 1 s

       d (1) = 2.6667 1²

       d (1) = 2.6667 feet

 t = 4 s

       d (4) = 2.6667 4²

       d (4) = 42.6672 feet

Let's calculate the speed

       v = Δx / Δt

       v = (42.6672 -2.6667) / (4-1)

       v = 40/3

        v = 13.33335 feet / s

        v = 13 feet / s

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Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an e
Oksi-84 [34.3K]

Answer:

0.69 ohm

Explanation:

Heat generated per second, H = 50 cal/s

Potential difference, V = 12 V

Let R is the resistance of coil.

The formula for the heat is given by

H = \frac{V^{2}}{R}t

50\times 4.186 = \frac{12^{2}}{R}\times 1

R = 0.69 ohm

3 0
2 years ago
A person is standing outdoors in the shade where the temperature is 35 °C.
Free_Kalibri [48]

Answer:

(a) Eₐ = 6.36 J/s

(b) Eₐ = 4.64 J/s

Explanation:

Stefan-Boltzmann law: States that the total energy per second radiated or absorbed by a black body is directly proportional to the absolute temperature.

Using, Stefan-Boltzmann equation

Eₐ =eσAT⁴ ................ Equation 1

where Eₐ = Radiant energy absorbed per seconds, e = emissivity, σ = stefan - boltzman constant, A = Surface area. and T = temperature in kelvin

(a) Where e = 0.89, σ = 5.67 ×10⁻⁸ watt/m²/K⁴, A = 140 cm² = 140 cm²(m²/10000cm²) = 0.014 m², T = 35 °C = (35 + 273) K = 308 K.

Applying these values in equation 1 above,

Eₐ = 0.89 × 5.67 ×10⁻⁸ × 0.014 × (308)⁴

Eₐ =6.36 J/s

(b) when e = 0.65,

∴ Eₐ = 0.65 × 5.67 × 10⁻⁸ × 0.014 × (308)⁴

 Eₐ = 4.64 J/s

6 0
2 years ago
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

\frac{y-x}{y}=\frac{6}{15}

15 y - 15 x = 6 y

9 y = 15 x

y=\frac{5}{3}x

Differentiate with respect to time.

\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}

As given, dx/dt = 4 ft/s

\frac{dy}{dt}=\frac{5}{3}\times 4

\frac{dy}{dt}=\frac{20}{3} ft/s

6 0
2 years ago
An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ. Find an expression for
VikaD [51]
We know that
g = LcosΘ 
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>
8 0
2 years ago
Read 2 more answers
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