(b) Find the magnitude of force F~2 that is acting on the block

(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .

Answer:

**Answer:**

Normal force=7.48 N

**Explanation:**

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a good mathematical representation of efficiency e of any heat engine? Where, Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem W: the amount of work done by the engine during one cycle or during some time specified in the problem A) e=QhW B) e=QcQh C) e=QcW D) e=WQh E) e=WQc

How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 JB) 5 JC) 50 JD) 1 JE) 10 J

Fuel cells have been developed that can generate a large amount of energy. For example, a hydrogen fuel cell works by combining hydrogen and oxygen gas to produce water and electrical energy. If a fuel cell can generate 10.0 kilowatts of power and the current is 15.8 amps, what is the voltage of the electricity?A. 0.63 voltsB. 158voltsC. 633 voltsD. 158,000 voltsE. 5.8 volts

Chuck wants to investigate how gas moleculesmove in a container. Which model would be mosthelpful to represent this motion?A. stacking blocks to build a towerB. freezing water in an ice cube trayC. bouncing elastic balls off of each other andthe walls of a roomD. placing a closed, water-filled plastic bag inthe sun and watching condensation form

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, with a standard deviation of σ = 0.5°C. How many readings must be taken so that the probability is 0.90 that the average is within ±0.1◦C of the actual temperature? Round the answer to the next largest whole number.

How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 JB) 5 JC) 50 JD) 1 JE) 10 J

Fuel cells have been developed that can generate a large amount of energy. For example, a hydrogen fuel cell works by combining hydrogen and oxygen gas to produce water and electrical energy. If a fuel cell can generate 10.0 kilowatts of power and the current is 15.8 amps, what is the voltage of the electricity?A. 0.63 voltsB. 158voltsC. 633 voltsD. 158,000 voltsE. 5.8 volts

Chuck wants to investigate how gas moleculesmove in a container. Which model would be mosthelpful to represent this motion?A. stacking blocks to build a towerB. freezing water in an ice cube trayC. bouncing elastic balls off of each other andthe walls of a roomD. placing a closed, water-filled plastic bag inthe sun and watching condensation form

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, with a standard deviation of σ = 0.5°C. How many readings must be taken so that the probability is 0.90 that the average is within ±0.1◦C of the actual temperature? Round the answer to the next largest whole number.

**Answer:**

force = 11.33

**Explanation:**

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

from newton second law of motion

Kgm/s^2

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**Answer:**

**Explanation:**

The system mosquito-raindrop is described by the Principle of Momentum Conservation:

The final speed is:

**Answer:**

Glycolysis

**Explanation:**

In human body, glucose levels are regulated by hormones insulin and glucagon, secreted from pancreas. Glucagon from alpha cells and insulin from beta cells of the pancreas. Glucagon are regulated along depending upon the blood sugar levels. During fasting when blood sugar levels are decreased the glucagon levels are increased. Glucagon increases hepatic glucose through glycogenelysis.

So, Glycolysis of glucagon is least likely to be activated by glucagon in hepatocytes

This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.

This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.

To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.

Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:

0.35*F2 = 3.25*F1 (1)

Because the net force should also be zero in equilibrium, we have:

F1 + F2 = 205.8 (2)

Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.

#SPJ12

To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a **force** of approximately 114N is required

The subject question is a classic example of **Torque** problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The **weight** of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.

This torque is calculated as Torque = r * F = 1.8m (**distance** from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.

#SPJ11

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final** velocity** of the combined carts to the initial velocity of the first cart.

To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.

After calculating the mass of the second cart, we can use the conservation of **momentum** again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.

#SPJ11

**Answer:**

A neap tide. Hope this helps

**Explanation:**