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zlopas [31]
3 years ago
5

A boat moves through the water with two forces acting on it. One is a 2.51×103 N forward push by the motor, and the other is a 1

.88×103 N resistive force due to the water. What is the acceleration of the 1384.1 kg boat? Answer in units of m/s 2
Physics
1 answer:
telo118 [61]3 years ago
7 0

Answer:

Acceleration a=0.455m/sec^2

Explanation:

We have given that the force by the motor F=2.51\times 10^3N=2510N

Resistive force by the water F=1.88\times 10^3N=1880N

So net force F_{NET}=2510-1880=630N

Mass of the boat = 1384.1 kg

According to newton's second law F=Ma, here M is mass and a is acceleration

So 630=1384.1\times a

a=0.455m/sec^2

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3 years ago
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A solar eclipse occurs when the Moon moves directly between the Earth and the Sun. When Venus moves directly between the Earth a
Serga [27]

Answer:

it's because the eclipse hides the moon and Venus and the sun

Explanation:

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7 0
2 years ago
1. A 1.32 x 104 meter steel railroad track with a coefficient of linear expansion of 12 x 10-6 per degree Celsius changes temper
GarryVolchara [31]

Answer:

1) 0.1584 m

2) To allow for expansion without derailment

3) 0.101376 m

4) 213.675 °C

5) 266.67 m

6) 8.33 × 10⁻⁶ /°C

7) The alloy meets the requirement

8) 1.95 × 10⁻³ /°C

9) 32.095 m

10) -12157.72°C

Explanation:

1) Equation for the coefficient of linear expansion = \frac{\Delta L}{L} = \alpha _L \Delta T

Where:

ΔL = Change in length = Required

L = Initial length = 1.32 × 10⁴ m

\alpha _L = Coefficient of linear expansion of steel = 12 × 10⁻⁶ /°C

ΔT = Change in temperature = 37°C - 27°C = 10°C

Plugging the values in the equation for the temperature expansion of steel, we have m;

ΔL = L × \alpha _L ×ΔT = 1.32 × 10⁴ × 12 × 10⁻⁶ × 10  = 0.1584 m

2. Here we have that by segmenting railroad tracks into short pieces, the expansion of the metal tracks with temperature can be absorbed by the gaps between the segment without distorting the shape and direction (pattern) of the tracks

3. Here we have;

\alpha _L = Coefficient of linear expansion of iron = 12 × 10⁻⁶ /°C

ΔT = Temperature change = 27°C - 3°C = 24°C

L = Height of the Eiffel Tower = 352 meters

∴ ΔL = L × \alpha _L ×ΔT = 352 × 12 × 10⁻⁶ × 24 = 0.101376 m

Therefore, the height of the Eiffel Tower changes from 352 m to about 352.101376 m each year, with an average change in height experienced each year = 0.101376 m

4. Here, we have

L = 13.0 ft

ΔL = 1 in.

\alpha _L = 30 × 10⁻⁶ /°C

ΔT = Required temperature change

From  \frac{\Delta L}{L} = \alpha _L \Delta T

\Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{1}{156 \times 30 \times 10^{-6}} = 213.675^{\circ}C

5. Here, we have;

L = \frac{\Delta L}{\alpha _L \Delta T}

∴ L = 1/(150×25 × 10⁻⁶) = 266.67 m

The bars original length = 266.67 m

6. Here we have;

\alpha _L = \frac{\Delta L}{L \times \Delta T}

Where:

ΔL = 3.00 - 3.002 = 0.002 m

L = 3.00 m

ΔT = 110°C - 30°C = 80°C

∴ \alpha _L = 0.002/(3.00 × 80) = 8.33 × 10⁻⁶ /°C

7. Here we have;

ΔL = L × \alpha _L ×ΔT = 3 × 8.33 × 10⁻⁶ × 210 = 0.00525 m

Therefore, final length = 3.00 m + 0.00525 m = 3.00525 m

Since 3.00525 m < 3.017 m hence the alloy meets the requirement.

8. Here, we have

L = 3.2 m

ΔL = 0.5 m

ΔT = 84°C - 24°C = 60°C

∴ \alpha _L = 0.5/(3.2 × 60) = 1.95 × 10⁻³ /°C

The coefficient of linear expansion of the material from which the rod is made = 1.95 × 10⁻³ /°C

9. Here, we have

Length of steel girder, L = 32.10 m

ΔT = 8°C - 22°C = -14°C

\alpha _L = 12 × 10⁻⁶ /°C

ΔL = L × \alpha _L ×ΔT

Hence ΔL = 32.1 × 12 × 10⁻⁶× -14 = -0.0054 m

New length = 32.1 - 0.0054 = 32.095 m

10. Here we have;

ΔL = 92.6 cm - 123 cm  = -30.4 cm

\alpha _L = 2.0 × 10⁻⁵ /°C

L = 123 cm

∴ \Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{-30.4}{123 \times 2.0 \times 10^{-5}} = -12357.724^{\circ}C

Therefore, the temperature will be 200 - 12357.724 = -12157.72°C.

3 0
3 years ago
A 1.0 kg object is attached to a string 0.50 m. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s. A b
aivan3 [116]
<span>50 N The centripetal force upon an object is expressed as F = mv^2/r So let's substitute the known values and calculate F = mv^2/r F = 1.0 kg * (5.0 m/s)^2 / 0.5 m F = 1.0 kg * 25 m^2/s^2 / 0.5 m F = 25 kg*m^2/s^2 / 0.5 m F = 50 kg*m/s^2 F = 50 N So the answer is 50 N which matches one of the available choices.</span>
6 0
3 years ago
A 500 kg rollercoaster car starts from rest at the top of a 10.0 m tall hill. it then travels down the track and up a loop. the
malfutka [58]

Speed of the roller coaster at the top of the loop= 7.67 m/s

Explanation:

using the law of conservation of energy

KEi + PEi= KEf + PEf

KEi= kinetic energy at the top of the hill=0 because the car is at rest there.

PEi= potential energy at the top of the hill

PEf= potential energy at the top of the loop

KEf= kinetic energy at the top of the loop

Also kinetic energy= 1/ 2m v² and potential energy= mgh

m= mass

h= height

v= velocity

so 0+ mghi = 1/2mv² + mg h

500 (9.8)(10)+ 1/2 (500) v²= 500 ( 9.8) (7)

49000+250 v²= 34300

250v²= 14700

v²=58.8

v=7.67 m/s

4 0
3 years ago
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