According to the U.S. Green Building Council, what percentage of the world’s energy use and greenhouse gas emissions can be attributed to buildings?

Answers

Answer 1
Answer:

According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions

Answer 2
Answer:

Answer:

40%

Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.


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What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking
controlled braking
threshold braking
coasting ​

Answers

Answer:

Controlled braking

Explanation:

CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.

Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.

Final answer:

The method of braking that involves applying smooth, steady pressure to the brake to bring the vehicle to a smooth stop is called controlled braking. It helps prevent skidding and provides the driver with more control over the vehicle.

Explanation:

The form of braking used to bring a vehicle to a smooth stop by applying smooth, steady pressure to the brake is known as controlled braking. This method of braking involves applying consistent, even pressure to the brake pedal, which allows the car to slow down gently and gradually. It helps prevent uncontrolled skidding and provides the driver with more control over the vehicle's direction and speed during the stop. Unlike other methods like trail braking, threshold braking, or coasting, controlled braking is typically the safest and most effective method for daily driving conditions.

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An internal explosion breaks an object, initially at rest,intotwo pieces, one of which has 1.5 times the mass of the other.If
7500 J were released in the explosion, how much kinetic energydid
each piece acquire?

Answers

Answer:

4500 J and 3000 J

Explanation:

According to conservation of momentum

      0 = m_1 V_1 + m_2 V_2

Given that m_2 = 1.5 m_1 , so

    V_1 = -1.5 V_2

  the kinetic energy of each piece is

    K_2= (1)/(2) m_2v_2^2

    K_1= (1)/(2) m_1v_1^2

substituting the value of V1 in the above equation

    K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2

  Given that

         K_1 + k_2 = 7500 J

       1.5 K_2 + K_2 = 7500

         K_2 = 7500 / 2.5

               = 3000 J

this is the KE of heavier mass

      K_1 = 7500 - 3000 = 4500 J

this is the KE of lighter mass

Final answer:

The question is about finding the kinetic energy acquired by each of two pieces of an object following an internal explosion, using principles of conservation of energy and momentum in physics.

Explanation:

The student has asked about an internal explosion that breaks an object into two pieces with different masses, releasing a certain amount of kinetic energy in the process. This question involves applying the principle of conservation of energy and momentum to find the kinetic energy acquired by each piece post-explosion.

Assuming piece 1 has a mass of m and piece 2 has a mass of 1.5m, the total mass of the system is 2.5m. Since 7500 J of energy was released in the explosion, to find the kinetic energy of each piece, we can use the fact that the total kinetic energy is equal to the energy released during the explosion. Let the kinetic energy of the smaller piece be K1 and of the larger piece be K2. Because the object was initially at rest and momentum must be conserved, the momenta of the two pieces must be equal and opposite. This relationship allows us to derive the ratio of the kinetic energies. We can solve for K1 and K2 proportionally. Finally, because the kinetic energy is a scalar quantity, adding the kinetic energies of the two pieces will equal the total energy released.

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Two bodies, one hot and the other cold kept in vacuum.what will happen to the tempreture of bodies after some time.

Answers

Hot body will lose heat from it, and that heat will goes out from it through radiation, so it's temperature will decrease after some time.

In same manner, cold body will take the heat, and it's temperature will increase

Hope this helps!

How many hours does earth take to complete one rotation?

Answers

Answer:

24 hours take earth to complete rotation

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force

Answers

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

Answers

The minimum angle that the  ladder make with the floor before it slips is 51.34 Degree.

Given data:

The weight of ladder is, W = 100 N.

The length of ladder is, L = 8.0 m.

The coefficient of static friction between ladder and floor is, \mu =0.40.

Apply the Newton' law in vertical direction to obtain the value of Normal Force (P) as,

N = mg

And force along the horizontal direction is,

F= \mu * N\n\nF = \mu * mg

Now, taking the torque along the either end of ladder as,

-mgcos \theta * (L)/(2)+Fsin \theta * L =0\n\nmgcos \theta * (L)/(2) = Fsin \theta * L

Solving as,

mgcos \theta * (L)/(2) = (\mu mg) * sin \theta * L\n\ncos \theta * (1)/(2) = (\mu) * sin \theta\n\ntan \theta = (1)/(2 * 0.40 )\n\n\theta = tan^(-1)(1/0.80)\n\n\theta = 51.34^(\circ)

Thus, we can conclude that the minimum angle that the  ladder make with the floor before it slips is 51.34 Degree.

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Answer:

The minimum angle is 51.34°

Explanation:

Given that,

Weight of ladder = 100 N

Length = 8.0 m

Coefficient of static friction = 0.40

We need to calculate the normal force

Using Newtons law in vertical direction

F_(y)=n-mg

N-mg=0

N=mg

We need to calculate the normal force

Using Newtons law in horizontal direction

F_(s)=f_(s)-P

f_(s)-P=0

f_(s)=P

P=\mu mg

We need to calculate the minimum angle

Using torque about the point A then

-mg\cos\theta* AB+P\sin\theta* AC=0

Put the value into the formula

mg\cos\theta*((L)/(2))=\mu mg\sin\theta* L

\cos\theta*(1)/(2)=\mu\sin\theta

(1)/(2)=\mu*\tan\theta

\theta=\tan^(-1)((1)/(2*0.40))

\theta=51.34^(\circ)

Hence, The minimum angle is 51.34°