A block weighing 35 N is resting on a steel table (us = 0.40).The minimum force to start this block moving is ___ N.

Answers

Answer 1

Answer:

The static friction force is a force that prevents motion of an object kept on a surface, due to the interaction of the surface of the object and the surface on which it is kept

The minimum force required to start the block moving is 14 N

Reason:

The known parameter are;

The weight of the block, W = 35 N

The static friction of the surface of the table surface, $\mu_s$ = 0.40

Required:

To find the minimum force to start the block moving

Solution:

The force required to start the block moving is given by the frictional force that needs to be surpassed

Force required, F = Force of friction

Force of friction = Normal reaction × Coefficient of static friction

On a flat surface, the normal reaction = The weight of the object

∴ Force of friction, $F_f$ = Weight of the object × Coefficient of static friction

Which gives;

$F_f = W * \mu_s$

$F_f = 35 \, N * 0.40 = 14 \, N$

Force required, F = Force of friction, $F_f$ = 14 N

The minimum force required to start the block moving, F = 14 N

Learn more about the static friction force here:

brainly.com/question/13877320

Answer 2

Answer:

Answer: 14

Explanation:

Force(normal)*coefficient

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Calculate the average power required to lift a 490 Newton object of vertical distance of 2 m in 10 seconds

Answers

   Power = (work done) / (time to do the work)

   Work = (force) x (distance)

If you want to lift a 490N object, you'll need (at least) 490N of force.
So, when you lift it 2 meters vertically, the work you do is

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Heather is doing an acid-base reaction. She has 25.00 mL of hydrochloric acid of unknown concentration in a flask. She wants to add a basic solution to the acid drop by drop, until all the acid has reacted. Which tool would be best for adding the basic solution dropwise and for most precisely measuring the volume of solution added?

a graduated cylinder
an eye dropper
a beaker
a buret

Answers

Answer: a buret

A graduated cylinder and a beaker cannot be used. This is because the basic solution needs to be added drop by drop. This is not possible using a graduated cylinder or a beaker because it is difficult to pour drop by drop from these. An eye dropper cannot be used because it does not measure the volume being added.

A buret would be the best choice because it precisely measures the amount being added and we can add the basic solution to the acid drop by drop.

A burette would be the best tool for the required use.
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In which of the following are action and reaction forces involved?A. when hitting a baseball with a bat
B. walking
C. swimming
D. all of the above

Answers

Answer:

D. All of the above

Explanation:

Newton's third law states that:

"when an object A exerts a force on an object B, object B exerts an equal and opposite force on object A. These forces are called action and reaction"

Let's analyze each situation:

A. when hitting a baseball with a bat --> we have two objects (the baseball and the bat), so we have a pair of forces: the action is the force exerted by the baseball on the bat, while the reaction is the force exerted by the bat on the baseball

B. walking --> in this case, the two objects we should consider are the man walking and the road: the action is the force exerted by the feet of the man on the road, while the reaction is the force exerted by the road on the feet of the man, which allows him walking

C. swimming --> in this case, the two objects we should consider are the man swimming and the water around him: the action is the force exerted by the body of the man on the water, while the reaction is the force exerted by the water on the body of the man, which allows him to swim.

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Two cars collide at an intersection. Car A, with a mass of 1900 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 17.0 m\s. As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 60.0degrees south of east from the point of impact.Part A WAS: How fast were the enmeshed cars moving just after the collision? I got 8.66 for velocity in part a which was CORRECT but i can't figure out PART B??...Part B:How fast was car A going just before the collision

Answers

Part A: The enmeshed cars were moving at a velocity of approximately 8.66 m/s just after the collision.

Part B: Car A was traveling at a velocity of approximately 8.55 m/s just before the collision.

How to compute the above velocities

To find the speed of car A just before the collision in Part B, you can use the principle of conservation of momentum.

The total momentum of the system before the collision should equal the total momentum after the collision. You already know the total momentum after the collision from Part A, and now you want to find the velocity of car A just before the collision.

Let's denote:

- v_A as the initial velocity of car A before the collision.

- v_B as the initial velocity of car B before the collision.

In Part A, you found that the enmeshed cars were moving at a velocity of 8.66 m/s at an angle of 60 degrees south of east. You can split this velocity into its eastward and southward components. The eastward component of this velocity is:

v_east = 8.66 m/s * cos(60 degrees)

Now, you can use the conservation of momentum to set up an equation:

Total initial momentum = Total final momentum

(mass_A * v_A) + (mass_B * v_B) = (mass_A + mass_B) * 8.66 m/s (the final velocity you found in Part A)

Plug in the known values:

(1900 kg * v_A) + (1500 kg * v_B) = (1900 kg + 1500 kg) * 8.66 m/s

Now, you can solve for v_A:

(1900 kg * v_A) + (1500 kg * v_B) = 3400 kg * 8.66 m/s

1900 kg * v_A = 3400 kg * 8.66 m/s - 1500 kg * v_B

v_A = (3400 kg * 8.66 m/s - 1500 kg * v_B) / 1900 kg

Now, plug in the values from Part A to find v_A:

v_A = (3400 kg * 8.66 m/s - 1500 kg * 8.66 m/s) / 1900 kg

v_A = (29244 kg*m/s - 12990 kg*m/s) / 1900 kg

v_A = 16254 kg*m/s / 1900 kg

v_A ≈ 8.55 m/s

So, car A was going at approximately 8.55 m/s just before the collision in Part B.

Learn more about velocity at:

brainly.com/question/25905661

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Answers

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well machine do not feel pain of course and us humans and people do feel pain of course but if we fell pain and get hurt it slows down the process but machines on the other hand can keep going until the life source runs out. if i helped plz make mine the best

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Answers

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