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yawa3891 [41]
3 years ago
14

An object has an initial velocity v and uniform acceleration a. If it covers a distance d, then its final velocity, v, is given

by the expression.
a. Does this equation give the correct unit of measurement of vf?
b. Is this a base unit or a derived unit?
c. Rewrite this equation to find the distance traveled in terms of initial velocity, final velocity, and acceleration.
Physics
1 answer:
viktelen [127]3 years ago
5 0

Answer:

.

Explanation:

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A driver dices off of a diving board 25 meter high. If he dives with a initial velocity of -1.2m/s and his acceleration is -9.8
kifflom [539]

ggggggggggggggggggggggggggggggg

3 0
4 years 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
4 years ago
One Newton is equivalent to<br> A. 1 kg/s2<br> B. 1 kg*m/s<br> C. 1 kg*m/s2<br> D. 1 kg/s
tiny-mole [99]

Answer:

B

Explanation:

7 0
3 years ago
In a nuclear experiment a proton with kinetic energy 3.0 MeV moves in a circular path in a uniformmagnetic field. What energy mu
dlinn [17]

Answer:

a)     K = 3 MeV   b)   K=  1.5 MeV

Explanation:

We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.

        F = q v x B

We can also write this equation based on the modules of the vectors

        F = qv B sin θ

With Newton's second law

       F = ma

       F = m v² / r

       q v B = m v² / r

       v = q B r / m

The kinetic energy is

       K = ½ m v²

Substituting

       K = ½ m (q B r/ m)²

       K = ½ B² r²  q² / m

       K = (½ B² R²)  q²/m

The amount in brackets does not change during the experiment

      K = A  q² / m

For the proton

     K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J

With this data we can find the amount we call A

    A = K  m/q²

    A = 4.8 10⁻¹³ 1.67 10⁻²⁷ /(1.6 10⁻¹⁹)²

    A = 3.13 10⁻²

With this value we can write the equation

    K = 3.13 10⁻²  q² / m

Alpha particle

    m = 4 uma = 4 1.66 10⁻²⁷ kg

   K = 3.13 10⁻² (2 1.6 10⁻¹⁹)² / 4.0 1.66 10⁻²⁷

   K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV

   K = 3 MeV

Deuteron

   K = 3.13 10⁻² (1.6 10⁻¹⁹)²/2 1.66 10⁻²⁷

   K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)

   K = 1.5 10⁶ eV

   K=  1.5 MeV

6 0
3 years ago
What services do plants provide? Select the three that apply.
Irina18 [472]

The answer to this is B, C, and D. hope this helped

6 0
3 years ago
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