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3 years ago

A train traveling at a constant speed covers a distance of 960 meters in 30 seconds. What is the train's speed?

Physics

2 answers:

madreJ [45]3 years ago

6 0

Answer:

32 m/sec

Explanation:

Given,

⇒ Distance that is covered= 960 meters

⇒ Time taken= 30 seconds

We know that,

⇒ $Speed=\frac{distance}{time taken}$

⇒ $Speed =\frac{960}{30}$

Therefore, speed = 32 m/sec.

balu736 [363]3 years ago

4 0

32 meters per second

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The work output of a machine divided by the work input is the ____ of the machine.

s344n2d4d5 [400]

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7 0

4 years ago

A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil

user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

4 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

3 years ago

A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo

Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, $\omega=0.47\ rev/s=2.95\ rad/s$

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

$I={m_1r^2}$

$I={52\times 3.9^2}$

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

$I_2=\dfrac{m_2r^2}{2}$

$I_2=\dfrac{118\times (3.9)^2}{2}$

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

$I=I_1+I_2$

$I=790.92+897.39$

I = 1688.31 kg-m²

The angular momentum of the system is :

$L=I\times \omega$

$L=1688.31 \times 2.95$

$L=4980.5\ kg-m^2/s$

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0

3 years ago

Help pleaseeeeeeeeeeeeee

kap26 [50]

The frictional force is in the opposite direction

7 0

3 years ago