A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force.True OR False


HELP ME!!!!!!¡!!!!

Answers

Answer 1
Answer: I believe the correct answer is true. A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force. Hope this answer the question.

Related Questions

An object is being acted upon by three forces and moves with a constant velocity. One force is 60.0 N along the x-axis, the second in 75.0 N along the y-axis. What is the magnitude of the third force?
Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 56 m long that spin at 13 rpm .A. At the tip of a blade, what is the speed?B. At the tip of a blade, what is the centripetal acceleration?C. A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .D. What force is required to keep a 10 mg water droplet moving in this circular arc?E. What is the ratio of this force to the weight of a droplet?
Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?
WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”picture attached
A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. what is the recoil speed of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally.

A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact?

Answers

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

Final answer:

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.

Explanation:

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.

Learn more about conservation of momentum here:

brainly.com/question/33316833

#SPJ11

Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?

Answers

The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

v = \omega r,

Where r indicates the radius and \omega the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

When a tennis ball is spun around in a circle on a string and the string breaks the tennis ballwill be pulled in a curved path away from the center because of Centrifugal force
True or false

Answers

Answer:

False

whenever the string breaks, the ball will follow the straight line tangential path

Explanation:

No, the ball will not follow a curved path after the string breaks. Since, the the direction of velocity is tangential to each point of the circular motion. Therefore, it changes at every point. This produces an acceleration in the circle called centripetal acceleration. There is also a tangential component of acceleration acting on the ball during this motion.

So, whenever the string breaks, the ball will follow the straight line tangential path. Hence, the given statement is false.

A parallel-plate capacitor is charged and then disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?

Answers

Answer:

U/U₀ = 2

(factor of 2 i.e U = 2U₀)

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected

Explanation:

Energy stored in a capacitor can be expressed as;

U = 0.5CV^2 = Q^2/2C

And

C = ε₀ A/d

Where

C = capacitance

V = potential difference

Q = charge

A = Area of plates

d = distance between plates

So

U = Q^2/2C = dQ^2/2ε₀ A

The initial energy of the capacitor at d = d₀ is

U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1

When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.

The final energy stored in the capacitor at d = 2d₀ is

U = 2d₀Q^2/2ε₀ A ...2

The factor U/U₀ can be derived by substituting equation 1 and 2

U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )

Simplifying we have;

U/U₀ = 2

U = 2U₀

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.

A boat can travel in still water at 56 m/s. If the boats sails directly across a river that flows at 126 m/s. What is the boats speed relative to the ground

Answers

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

Calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Answers

Answer:

For 25-turn electromagnet, Number of clips = 4.1

For 50-turn electromagnet number of clips = 9.6

Explanation:

To calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Hence;

Using the equations gotten from the graph in the previous question and 1.0 V as the value for x, we get

For 25-turn electromagnet y = 3.663x * 0.5

(rounded to one decimal place) Number of clips = 4.1

For 50-turn electromagnet y = 7.133x 2.5

(rounded to one decimal place) Number of clips = 9.6