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fenix001 [56]
4 years ago
14

Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. T

he length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 14.1 ∘C above room temperature?
Physics
1 answer:
Deffense [45]4 years ago
4 0

Answer:

0.00034 m

Explanation:

Since the length of the aluminium bar, L is given by , L = 1.0000 + 2.4 × 10⁻⁵T and T = 14.1°C, we substitute the value of T into L. So, we have L = 1.0000 + 2.4 × 10⁻⁵ × 14.1°C = 1.0000 + 0.0003384 = 1.0003384 m. The change in length is thus 1.0003384 - 1.0000 = 0.0003384 m ≅ 0.00034 m

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An object of mass 9.00 kg attached to an ideal massless spring is pulled with a steady horizontal force across a frictionless le
frosja888 [35]

Answer:

Acceleration of the object will be 1.05m/sec^2

Explanation:

We have given mass of the object m = 9 kg

spring constant k = 95 N/m

Spring is stretched by 10 cm

So x = 10 cm = 0.1 m

We know that force is given by F=kx=95\times 0.1=9.5N

From newton's law we also know that force is given by

F = ma , here m is mass and a is acceleration

So 9.5=9\times a

a=1.05m/sec^2

5 0
4 years ago
Calculate the efficiency of a hair dryer which takes in 4500J of energy per second and transfers 150J as useful heat energy.
Basile [38]

Answer:

η = 0.0333 = 3.33 %

Explanation:

The efficiency of any device is defined as the ratio of the useful output obtained from that device to the input required by the device. Therefore, the general formula to find the efficiency of any device is given as follows:

\eta = \frac{O}{I}

where,

η = efficiency of hair dryer = ?

O = Useful Output Given by the Hair Dryer = 150 J

I = Input Energy Required by the Hair Dryer = 4500 J

Therefore,

\eta = \frac{150\ J}{4500\ J}\\

<u>η = 0.0333 = 3.33 %</u>

5 0
3 years ago
Alex goes cruising on his dirt bike. He rides 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south, 300 m east, and fi
velikii [3]

ANSWER

0\operatorname{km}

EXPLANATION

First, let us make a sketch of the question:

From the diagram:

=> black line represents North

=> green line represents East

=> blue line represents West

=> red line represents South

From the diagram, we see that at the end of his journey, he returns to his start point.

Since displacement is the measure of the change in the position of an object and the boy's position did not change after the journey, his displacement is:

0\operatorname{km}

7 0
1 year ago
If the balloon can barely lift an additional 3500 N of passengers, breakfast, and champagne when the outside air density is 1.23
Bingel [31]

The complete question is :

A hot-air balloon has a volume of 2100 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1300 N .

If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?

Solution :

Given volume of the hot air balloon $=2100 \ m^3$

The balloon fabric weights = 860 N

The weight of the basket with the gear and propane tank = 1300 N

Density of outside air $= 1.23 \ kg/m^3$

∴  Total pay load = Weight of the air displaced - weight of gas inside the balloon

Total pay load = 860 + 1300 + 3400

                        = 5560 N

Mass = density x volume

Weight  $= \text{mass} \times g$

Weight = volume x density $\times \text{ acceleration due to gravity (g)}$

Weight of the displaced air = 2100 x 1.23 x 9.8

                                             = 25313 N

Weight of the gas inside the balloon = density $\times \text{ acceleration due to gravity (g)}$ x volume

                                                             = density x 9.8 x 2100

                                                             = density x 20580 N

Therefore substituting the values, we get

⇒ 25313 - (density x 20580) = 5560

⇒ density $=\frac{19753}{20580}$

                 $= 0.96 \ kg/m^3$

So the density of the heated gas $= 0.96 \ kg/m^3$

8 0
3 years ago
The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875
Natali [406]

Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

F_m=m.a\\F_m=0.875kg*a

The acceleration is given by:

a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2

So the force exerted  by the motor is:

F_m=0.875kg*31.9m/s^2\\F_m=27.9N

The work done by the motor is given by:

W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m

W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J

And finally, the power is given by:

P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W

8 0
3 years ago
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