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3 years ago

In the northern hemisphere planetary winds deflect to the

Physics

1 answer:

umka2103 [35]3 years ago

8 0

Answer:

Planetary are deflected to right due to Coriolis effect.

Explanation:

The term Coriolis effect is defined as an effect in which the rotating object experience a force called as coriolis force which acts perpendicular to the axis of rotation and direction of motion. The effect talks about how the moving objects like ocean currents, wind are deflected due to the rotation of earth.

Winds and ocean currents are strongly affect by this effect.

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Which of the following group 7A elements is the most reactive?

nevsk [136]

Among the elements in group 7A, the most reactive would be Fluorine. Fluorine is a major component of toothpaste, what we use to brush and clean our teeth with. It is so reactive that it cannot be contained in glass.

8 0

4 years ago

Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr

lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time of computer A is $CT_{A} =200 ps$

Clock time of computer B is $CT_{B} =250 ps$

Effective CPI of computer A is $CPI_{A} =1.5$

Effective CPI of computer B is $CPI_{B} =1.7$

CPU time of A is

$CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec$

CPU time of B is

$CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec$

Hence Computer A is Faster by $\frac{425}{300} =1.41$

Computer A is 1.41 times faster than the Computer B

4 0

3 years ago

A 1.0-kg standard cart collides on a low-friction track with cart A. The standard cart has an initial x component of velocity of

Andru [333]

Answer:

Explanation:

Mathematically, linear momentum is expressed as the product of mass and velocity. Linear momentum conservation law states that a body or system of bodies retains its total momentum unless an external force is applied to the system.

In this case, the system consists of two carts.

At the start, the linear momentum (P) of the system is equal to:

$P=1.0kg*0.4m/s=0.4kg*m/s$

It's only composed of linear momentum of the standard cart because cart A doesn't have any linear momentum at that moment.

After the collision, linear momentum has to be the same

$P=0.4kg*m/s=1.0kg*0.20m/s+m_{A} *0.70m/s$

where m_A is the mass of the cart A.

Solving for m_A

$m_{A} =0.28kg$

After the cart A rebounds, the linea momentum of the system has changed (because of the force present in the rebound). The new linear momentum is:

$P=1kg*0.2m/s+0.29kg*(-0.7m/s)=-0.003kg*m/s$

Then, the lump of putty is added to the system, but the linear momentum has to be the same, because we added a mass, not a force. The mass of that putty (m_p) has to be added to the equation of the system

$-0.003kg*m/s=1.0kg*(-0.2m/s)+(0.29kg+m_{p} )(0.4m/s)$