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denis23 [38]
2 years ago
6

The longest passenger liner ever built was the France, 66,348 tons and 315.5 long. Suppose its front end passes the edge of a pi

er at a speed of 2.10 m/s while the ship is accelerating uniformly at 0.03 m/s^2. At what speed will the back end of the vessel pass the pier?
Physics
1 answer:
Kaylis [27]2 years ago
5 0

Answer:

The back end of the vessel will pass the pier at 4.83 m/s

Explanation:

This is purely a kinetics question (assuming we're ignoring drag and other forces) so the weight of the boat doesn't matter. We're given:

Δd = 315.5 m

vi = 2.10 m/s

a = 0.03 m/s^2

vf = ?

The kinetics equation that incorporates all these variables is:

vf^2 = vi^2 + 2aΔd

vf = √(2.1^2 + 2(0.03)(315.5))

vf = 4.83 m/s

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With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the c
Alinara [238K]

Answer:

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

Explanation:

An object's weigh due to the gravitational attraction force of the earth is:

w = mg

            Where: m is the object's mass

                         g is the  gravitational acceleration in the surface earth

                         g = 9.8 m/s2

The the ultralight car's weight is:

w_{uc} = (540)(9.8)

w_{uc} = 5292 N

And the Honda Accord's weight is:

w_{HA} = (1450)(9.8)

w_{HA} = 14210 N

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

4 0
3 years ago
PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)
denpristay [2]

Answer:

False, true

Explanation:

3 0
3 years ago
Read 2 more answers
Multiply 0.00032 cm by 4.02 cm and express the answer in scientific notation
FromTheMoon [43]
0.00032cm*4.02=1.2864 × 10^-3 in scientific notation.
4 0
3 years ago
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A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
3 years ago
Q = cmAT
castortr0y [4]

Answer: Q=3000 cal

Explanation:

We are given the following formula:

Q=m. c. \Delta T   (1)

Where:

Q=3000 cal is the amount of heat

m=300g  is the mass  of water

c=1 cal/g \°C  is the specific heat of water

\Delta T  is the variation in temperature, which in this case is  \Delta T=30\°C-20\°C=10\°C  

Rewriting equation (1) with the known values at the right side, we will prove the result is 3000 cal:

Q=(300g)(1 cal/g \°C)(10\°C)   (2)

Q=3000 cal   This is the result

8 0
2 years ago
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