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tensa zangetsu [6.8K]
2 years ago
7

Which of the statements given below is correct? Give reason

Physics
1 answer:
trapecia [35]2 years ago
7 0
I’m the summer the winds flow from land towards the ocean
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6. traction a. friction between a tire and the road.b. pushes a moving object out of a curve and into a straight linec. the abil
Gennadij [26K]

Answer:

6. a. friction between a tire and the road

7. c. energy of motion

8. c. the force with which a moving vehicle hits another object

Explanation:

6. As a car moves along the road, the tires push back against the ground. As tires push back against the ground, the road exerts and opposing force to the motion of the tires. This opposing force is the friction between the tires and the road. <u>This opposing force between the tires and the rad is called traction.</u>

So, the answer is a

7. As an object moves, it has energy. <u>This energy due to its motion is called kinetic energy.</u>

So, the answer is c

8. When a moving vehicle hits another object, it exerts a force on the object. The process of the vehicle hitting the other object is called impact and the<u> force exerted on the object is called the force of impact. </u>

So, the answer is c.

4 0
3 years ago
​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
vlada-n [284]

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0
3 years ago
Which of these letters is the symbol for current in equations A: c B: i C: r D: t
True [87]
  Answer:  [B]:  The letter, "<em /> I " ; for current;  in units of "Amps" .
__________________________________________________
4 0
3 years ago
A large box has a mass of 500kg and the coefficient of static friction for the box and the floor is 0.45, and the coefficient of
Mumz [18]
A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.

b) The acceleration will be the horizontal force - the kinetic friction force (since they act in opposite directions) divided by the mass. Kinetic friction force = (coefficient of kinetic friction)*(normal force or weight). 

F(net) = (2207.25 N)-(0.30)(4905 N) = 735.75 N

a = (735.75 N)/(500kg)= 1.4715 m/s^2
4 0
3 years ago
Consider the following thermochemical equation:
Arisa [49]

Answer:

m_w=5.3248\times 10^{23}\ kg

Explanation:

According to the given reaction,

heat released by thecombustion of 2 molecules of Methanol, \Delta H=1277000\ J

we know that molecular mass of Methanol, m=32\ g.mol^{-1}

∴12 gram of methanol = \frac{3}{8}\ moles

we know 1 mole = 6.023\times 10^{23}\ molecules

so,

\rm \frac{3}{8} \ moles=2.258\times 10^{23}\ molecules

Heat from the combustion of 2.258\times 10^{23}\ molecules:

Q=1277000\times \frac{2.258\times 10^{23}}{2}

Q=1442132.0625\times 10^{23}\ J

<u>Now the  mass of water that can be heated from 23.5°C to 88.2°C :</u>

Q=m_w.c_w.\Delta T

1442132.0625\times 10^{23}=m_w.4186\times (88.2-23.5)

m_w=5.3248\times 10^{23}\ kg

8 0
3 years ago
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