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kompoz [17]
3 years ago
6

Convert as indicated step by step 1)1 sec into day 2) 1 week into sec (3) 50cm into meter​

Physics
1 answer:
Allushta [10]3 years ago
4 0

Answer:

1) 1,157*10⁻⁵ day; 2) 604800 sec.; 3) 0.5 m.

Explanation:

1) 1 second into one day:

one day consists of 24 hours, it means 24*3600=86400 seconds. Then 1 second is 1/86400 day or 1,157*10⁻⁵ day.

2) 1 week into seconds:

1 day consists of 3600*24=86400 seconds, then the week is 7*86400=604800 seconds.

3) 50 cm into meter:

100 cm is one meter, then 50 cm is 0.5 meter.

PS. note, there is not the only way to perform conversion.

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A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
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Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
2 years ago
A student uses the right-hand rule as shown.
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Answer: O:right

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2 years ago
A group of atoms with aligned magnetic poles are known as which of the following?
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4 0
3 years ago
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A car runs for about 10 years during its average useful life. If the odometer reads 120,000 miles at the end, what was the avera
Eva8 [605]

Answer:

12000 mph

Explanation:

Given that,

Distance read by the odometer = 120,000 miles

Duration of car, t = 10 years

We need to find the average speed of the car. Speed of an object is equal to the total distance covered divided by total time taken. So,

v=\dfrac{120000}{10}\\\\=12000\ mph

So, the average speed of the car is 12000 mph.

7 0
3 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0
3 years ago
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