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3 years ago

12

It took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 km/h. How far did the train travel

between the two cities?

1 answer:

Nadusha1986 [10]3 years ago

6 0

Answer: 420 km

Explanation:

120 per hour, 3.5 hours

120 x 3.5 = 420

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Shkiper50 [21]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg

anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

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3 years ago

A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 130-kg laser sensor that measures the thi

umka21 [38]

Answer:T=1316.21 N

Explanation:

The tension has two components: Vertical and Horizontal. The

horizontal component is ma, the vertical component is mg. Using

Pythagoras theorem, we can find the tension as:

T=((ma)^2 (mg)^2)^(1/2)

So

T=((129*2.84)^2 (129*9.8)^2)^(1/2)

T=1316.21 N

8 0

2 years ago

Could I get help on this question please . My parents won’t help me /:

vovikov84 [41]

Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

From the question given above, the following data were obtained:

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

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3 years ago

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un

Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within 1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

$B_{i} =\frac{hr/2}{k}$

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

$\pi$ Dh(T-Tinf)= $I^{2} R_{e}$

make T the subject of the equation , we have

T=25+ $\frac{100^2*0.01}{\pi*0.001*500 }$

T=88.7 degC

rate of chnage in temperature

dT/dt= $\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)$

at t=o and integrating both sides $\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}$

we have

$\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}$

t=8.31s

steady state temperature =88.7deg C

t=time within 1 degC of it steady stae is 8.31s

7 0

3 years ago