How much force is required to accelerate a 2 kg mass at 3 m/s2

Answers

Answer 1
Answer:
Force = (mass) x (acceleration)                  Newton's second law of motion.

Force = (2 kg) x (3 m/s²)  =  6 newtons.


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Calculate the gravitational potential energy of a 10kg box on the top shelf of a closet, 2m above the floor

Answers

Answer:

196 J

Explanation:

PE = mgh

PE = (10 kg) (9.8 m/s²) (2 m)

PE = 196 J

Convert an absolute pressure of 6.31 ✕ 105 N/m2 to gauge pressure in lb/in2

Answers

Answer:

The gauge pressure is = 91.52 lb/in²

Explanation:

Given data,

The value of pressure in S.I unit, = 6.31 x 10⁵ N/m²

The conversion formula for SI unit into US Customary Units is given by,

                             1 N/m² = 1.45 x 10⁻⁴ lb/in²

Using the conversion formula, the new value of pressure becomes,

                        Pressure = 6.31 x 10⁵ X 1.45 x 10⁻⁴ lb/in²

                                         = 91.52 lb/in²

Hence, the gauge pressure is, ς = 91.52 lb/in²

Which statement about speed and/or velocity is true?Velocity has both speed and direction.

Speed is constant.

Speed has direction and velocity.

Speed is measured over time.

Answers

The statement about speed and/or velocity which is true is: Velocity has both speed and direction.

Velocity refers to the rate of change in displacement (distance) with time. Velocity is a vector quantity and this simply means that, it has both magnitude (speed) and direction.

Mathematically, velocity is given by the formula;

Velocity = (Distance)/(Time)

On a related note, speed is a scalar quantity because it has magnitude but no direction.

However, both speed and velocity are measured in meters per seconds.

In conclusion, the relationship between speed and velocity is that velocity has both speed and direction.

Read more: brainly.com/question/17742679

Answer:

Velocity has both speed and direction.

Explanation:

speed is a part of velocity

A stress is considered a "bad" stress when it A. cannot be eliminated. B. keeps a person from feeling good. C. challenges a person to do better. D. doesn't lead to a productive end.

Answers

The answer is either c or d

Calculate the angular velocity of a clocks second hand, its minute hand, and its hour hand. State in Rad/s. What is its angular acceleration in each case?

Answers

The angular velocity of a clocks second hand, its minute hand, and its hour hand are 0.1047rad/s, 1.745 × 10⁻³rad/s and 1.454 × 10⁻⁴rad/s respectively.

Time period for second hand; T = 60s

Time period for minute hand; T = 60m = 60 * 60s = 3600s

Time period for hour hand; T = 12hours = 12* 60 * 60s = 43200s

Now, we use the relation between angular speed and time period:

w = (2\pi )/(T)

Where ω is the angular velocity and T is the time period in seconds.

For Second hand

w = (2\pi )/(60s) = 0.1047rad/s

For Minute hand

w = (2\pi )/(3600s) = 1.745* 10^(-3)rad/s

For Hour hand

w = (2\pi )/(43200s) = 1.454* 10^(-4)rad/s

Therefore, the angular velocity of a clocks second hand, its minute hand, and its hour hand are 0.1047rad/s, 1.745 × 10⁻³rad/s and 1.454 × 10⁻⁴rad/s respectively.

Learn more: brainly.com/question/8711708

Second hand:

1 rev per minute = (2π radians/minute) x (1 min/60sec) = π/30 rad/sec

Minute hand:

1 rev per hour = (2π radians/hour) x (1 hr/3600 sec) = π/1800 rad/sec

Hour hand:

1 rev per 12 hours = (2π rad/12 hr) x (1 hr/3600 sec) = π/21,600 rad/sec

As long as the clock is in good working order, and the hands are turning steadily at their normal rate, there is no angular acceleration.

What kind of image is formed by a convex lens when the object is less than a focal length away from the object?A) an image that is always smaller than the object

B) none, because the lens is too close to the object

C) a virtual image

D) an inverted real image

Answers

Answer:

C

Explanation:

An object that is less than a focal length from a convex lens forms a virtual, magnified image. When the object is closer to a convex lens than the focal point, its rays do not pass through the focal point to form an image on the opposite side of the lens.