2560 meters in 60 seconds, what is his average speed (velocity)

An apple with a mass of 0.95 kilograms hangs from 3.0 meters above the ground. What is the potential energy of the apple

Vesnalui [34]

As Potential energy =mgh

m= 0.95kg

h=3 meter

g = 9.8 m/sec^2. ( acceleration due to gravity)

So P.E =(0.95)(9.8)(3)kgm^2/s^2

P.E =27.93 joules

3 0

2 years ago

The variable that you change during an experiment is called what type of variable?

elena-s [515]

The independent variable

7 0

3 years ago

Read 2 more answers

A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i

Stolb23 [73]

The speed $V_{i}$ of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.

We are given that-

the mass of the elevator (m) = 1000 kg ;

the distance the elevator decelerated to be y = 8m ;

the tension is T = 11000 N;

let us determine the acceleration 'a' by using Newton's second law of motion.

∑Fy = ma

W - T = ma

(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a

9800 - 11000 = 1000

a = - 1.2 m/s²

Using the equation of kinematics to determine the initial velocity.

$V_{f}$ ² = $V_{i}$ ² + 2ay

$V_{i}$ = √ ( 2 x 1.2m/s² x 8 m )

$V_{i}$ = √19.2 m²/s²

$V_{i}$ = 4.38 m/s ≈ 4 m/s

Hence, the initial velocity of the elevator is 4m/s.

Read more about the Equation of kinematics:

brainly.com/question/12351668

#SPJ4

8 0

1 year ago

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi

True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0

3 years ago

2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n

AleksandrR [38]

Answer:

No

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

$d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}$

For the ships to sight each other, distance must be 5 or smaller

$d \leq 5$

$\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5$

$(12 - 12t)^2 + (9t)^2 \leq 25$

$144t^2 - 288t + 144 + 81t^2 - 25 \leq 0$

$225t^2 - 288t + 119 \leq 0$

$(15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0$

$(15t^2 - 9.6)^2 + 26.84 \leq 0$

Since $(15t^2 - 9.6)^2 \geq 0$ then

$(15t^2 - 9.6)^2 + 26.84 > 0$

So our equation has no solution, the answer is no, the 2 ships never sight each other.

8 0

3 years ago