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aalyn [17]
3 years ago
15

Superman stops a "speeding locomotive" of mass 8.0x103 kg in 4.0 seconds. If the train was originally moving at 40.m/s,

Physics
1 answer:
madreJ [45]3 years ago
8 0

Hello!

<em>a) What is the change in momentum</em>?

The change of the momentum is the velocity, because the velocity is reduced to zero.

We can calculate the aceleration, applicating the formula:

\boxed{a=\frac{V-Vi}{t} }

Like you see, the final velocity will be zero, so:

a = \dfrac{0m/s-40m/s}{4s}

a = \dfrac{-40m/s}{4s}

a = -10\ m/s^{2}

Like you see, the aceleration applicate by superman is of <u>-10 m/s^2.</u>

If the aceleration is negattive, means the velocity will decrease.

<em>b.) What is the magnitud of the force wich superman exerts on the train?</em>

For calculate the force applicate for superman, lets applicate the second law of Newton:

\boxed{F=ma}

Lets replace and resolve it:

F = 8x10^3 kg * (-10 m/s^2)

F = 8000 kg * (-10 m/s^2)

F = -80 000 N

The force applicated is of <u>-80000 Newtons.</u>

When the force is negative means the force applicated in the another direction of the object what is traveling.

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Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478
klasskru [66]

Answer:

a) 2.5 m/s²

b) 6.12 m/s

Explanation:

Tension of rope = T = 356N

Weight of material = W = 478 N

Distance from the ground = s = 7.5 m

Acceleration due to gravity = g = 9.81 m/s²

Mass of material = m = 478/9.81 = 48.72

Final velocity before the bundle hits the ground = v

Initial velocity = u = 0

Acceleration experienced by the material when being lowered = a

a) W-T = ma

⇒478-356 = 48.72×a

\Rightarrow \frac{122}{48.72} = a

⇒a = 2.5 m/s²

∴ Acceleration achieved by the material is 2.5 m/s²

b) v²-u² = 2as

⇒v²-0 = 2×2.5×7.5

⇒v² = 37.5

⇒v = 6.12 m/s

∴ Velocity of the material before hitting the ground is 6.12 m/s

5 0
3 years ago
Calculate the kinetic energy, in joules of a 1160-kg automobile moving at 19.0 m/s.
Rudiy27

Answer:

Explanation:

K.E=1/2 mv²

m=1160 kg

v=19.0 m/s

so k.e=1/2*1160*(19.0)²

K.E=1/2*1160*361

K.E=1/2*418760

K.E=209380=2.0*10^5 j

7 0
3 years ago
Select the correct answer
Nadya [2.5K]

Answer:

f= 3.0 \times 10 {}^{8} \div 7.0 \times 10 {}^{7} \\ f = 4.28hz

Given

7 0
3 years ago
A converging lens of focal length 20 cm is placed in contact with a diverging lens of focal length 30 cm. The focal length of th
kramer

Answer:

The focal lenth (F) =+10.0cm

Explanation:

The formular for combined focal length (F) is given as;

1/f = 1/f_{1} + 1/f_{2}

In this  question,

F1 = 20cm

F2  = -30cm

Plugging the values into the formuar above,

1/f = 1/20 + 1/(-30)

1/f = 0.05 - 0.033[tex]1/f = -0.017f = [tex]1/ -0.017

f = 58.82cm

i.e. the combination behaves as a converging lens (because of the postive sign) of focal length 58.82cm .

7 0
3 years ago
A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i
padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

v=\sqrt{\dfrac{T}{\mu}}

Where, \mu = mass per unit length

T = tension

Put the value into the formula

v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}

v =519.61\ m/s

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0
3 years ago
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