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3 years ago

How fast must a train accelerates from rest to cover 518 m in the first 7.48 s?

Physics

1 answer:

Rus_ich [418]3 years ago

8 0

Heya!

For this problem, use the formula:

s = Vo * t + (at^2) / 2

Since the initial velocity is zero, the formula simplifies like this:

s = (at^2) / 2

Clear a:

2s = at^2

(2s) / t^2 = a

a = (2s) / t^2

Data:

s = Distance = 518 m

t = Time = 7,48 s

a = Aceleration = ¿?

Replace according formula:

a = (2*518 m) / (7,48 s)^2

Resolving:

a = 1036 m / 55,95 s^2

a = 23,34 m/s^2

The aceleration must be <u>23,34 meters per second squared</u>

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Assume you need to design a hydronic system that can deliver 80,000 Btu/hr. What flow rate of water is required if the temperatu

PolarNik [594]

Answer:

At 10°F change in temperature

Mass flowrate = 1.01 kg/s = 2.227 lbm/s

Volumetric flowrate = 1010 m³/s = 35667.8 ft³/s

At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

ṁ = 23445.7/(4182 × 11.111)

ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

4 0

3 years ago

ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

8 0

2 years ago

What's the value of 57,281 joules in Btu? A. 54.3 Btu B. 14.2 Btu C. 28.9 Btu D. 37.7 Btu

Alex Ar [27]

I'm quite sure it is A

8 0

3 years ago

Read 2 more answers

Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .

lara [203]

Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

initial speed of the of the rock, v = 17 m/s

time taken by the Heather rock to reach water

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 17 t - 0.049 = 0

now,

$t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² - 17 t - 0.049 = 0

now,

$t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₂ = 3.47 s ,-0.0028 s

rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

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2 years ago

If there are oncoming cars to the left of you and a child on a bike to the right, what is the correct action to take

tankabanditka [31]

Answer:

Give the child a lot of room to your side, which may mean moving closer to the oncoming vehicles.

Explanation:

I majored in Physics.

3 0

3 years ago