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Artemon [7]
3 years ago
7

A powerful missile reaches a speed of 5 kilometers per second in 10 seconds after its launch. What is the average acceleration o

f the missile during this period?
A.
0.2 meters/second2
B.
0.4 centimeters/second2
C.
0.5 meters/second2
D.
0.5 kilometers/second2
Physics
1 answer:
bixtya [17]3 years ago
3 0

Answer:

D

Explanation:

5 Km=5000m

so Δv=5000 m/sec

a=Δv/Δt

=5000/10

a=500 m/sec²                 as 500÷1000=0.5 Km

a=0.5 km/sec²

so D is the right answer.

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How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{9}{40}

f=4.44\ cm

When , R₁= -4, R₂ = 8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{3}{40}

f=-13.33\ cm

When , R₁= 4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{3}{40}

f=13.33\ cm

When , R₁= -4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{9}{40}

f=-4.44\ cm

Hence, The lenses with different focal length are four.

8 0
3 years ago
Producers,____________, and_______________ help to move matter and energy through ecosystems.
Phoenix [80]
Producers, consumers, and decomposers help to move matter and energy through ecosystems.

Hope this helps! :)
3 0
3 years ago
an object moving with uniform acceleration has a velocity of 12.ocm/s. if its x coordinate 2.00 later is 25.00cm what is its acc
Viktor [21]
<span> We're given that x=25 when t=2: </span>

<span>25 = 3 + 12(2) + (1/2)a(2)^2 </span>

<span>Thus a = -1 cm/sec^2</span>
4 0
3 years ago
How much time would it take for the sound of thunder to travel 1500 meters if sound travels at a speed of 330 m/s
Elis [28]

Data given:

Δx=1500m

v=330m/s

t=?

Formula:

V=Δx/t

Solution:

t=1500m/330m/s

t=4.5s

7 0
3 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
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