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Elanso [62]
3 years ago
12

A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

speed of 0.76 m/s when it leaves the chute. Determine the speed of the car–coal system after the coal has come to rest in the car.
Physics
1 answer:
Nady [450]3 years ago
8 0

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

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A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0
3 years ago
On a violin, the highest note Mandy can play is an A-note, which produces a sound wave with a high frequency. The lowest note sh
lora16 [44]
I think the answer would be: The G-note's wavelength is longer

Here are the formula to calculate wavelength

Wavelength = Wave speed/Frequency

Which indicates that the wavelength will become larger as the frequency became smaller.

6 0
3 years ago
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andreev551 [17]

Answer:

0.438kg/ms-¹

Explanation:

Momentum, denoted by p, can be calculated by using the formula;

p = mv

Where;

m = mass (kg)

v = velocity (m/s)

Momentum (p) of bird = 0.216 kg × 5.87 m/s = 1.268kg/ms-¹

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Having calculated the momentum of the bird to be 1.268kg/ms-¹, and the momentum of the baby to be 1.706kg/ms-¹, the difference in momentum between the flying bird and the crawling baby is:

{1.706kg/ms-¹ - 1.268kg/ms-¹} = 0.438kg/ms-¹

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Why centre of mass equal to centre of gravity
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Because mass and distance determine gravity, so the more mass you have, the more gravity.
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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a
Vesna [10]

Answer:

Tangential speed=5.4 m/s

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Explanation:

We are given that

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We know that

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By using \pi=3.14

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Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

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