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koban [17]
3 years ago
8

a car has a velocity of 10ms-1. it accelerates at 0.2 ms-2 for half minute find the total distance travelled and final velocity

of car.​
Physics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

Final velocity = 16 m/s

Total distane travelled = 390 m

Explanation:

We can use equation of motion to solve this:

v = u + at \\ v = 10 + 0.2(30) \\ v = 16 {ms}^{ - 1}

s = ut +  \frac{1}{2} a {t}^{2}  \\ s = 10(30) +  \frac{1}{2} (0.2) {(30)}^{2}  \\ s = 390m

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A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe
liubo4ka [24]

Answer:

 ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

I_1=\dfrac{M+m}{2}r^2

I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2

I_1=0.75\ kg.m^2

Final mass moment of inertia

I_2=\dfrac{M}{2}r^2

I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2

I_2=0.468\ kg.m^2

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s

ω₂=1.20

7 0
3 years ago
A symbolic model for learning is a model that is observed in person.
kap26 [50]

Answer:

True

Explanation:

3 0
3 years ago
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Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On
8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.

4 0
3 years ago
Please help!!
Luba_88 [7]

Answer:

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3 years ago
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A 0.75 kg book is pushed across the table with an acceleration of 0.3 m/s2. What force is being applied to the
Brums [2.3K]

Answer:

\boxed {\boxed {\sf 0.225 \ Newtons}}

Explanation:

We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F= ma

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

  • m= 0.75 kg
  • a= 0.3 m/s²

F= 0.75 \ kg * 0.3 \ m/s^2

Multiply.

F =0.225 \ kg * m/s^2

1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.

F= 0.225  \ N

<u>0.225 Newtons of force</u> are applied to the book.

5 0
3 years ago
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