How many seconds are there in 27.5 days?

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

2 years ago

Bodies weighing 1 kilogram and 5 kilograms lie on a smooth horizontal surface. If a traction force of 0.6 N acts on another 5 kg

natima [27]

0.6/5,1,5

so calculate it

not so sure though

6 0

3 years ago

An 4.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, starting

mina [271]

Answer:

2.64N

Explanation:

Force = mass * acceleration

Given

mass = 4kg

distance = 1.9m

Time t = 2.4s

Get the acceleration using the equation of motion

S = ut + 1/2at²

1.9 = 0 + 1/2a(2.4)²

1.9 = 5.76a/2

1.9 = 2.88a

a = 1.9/2.88

a = 0.66m/s²

Get the magnitude of the force

Force = 4 * 0.66

Force = 2.64N

Hence the net force acting on the fish is 2.64N

5 0

2 years ago

A vacuum cleaner has a rating of 460 W on 230 V mains. The value of the fuse connected in the plug will be

amid [387]

Answer:

2 amps

Explanation:

Given data

Power = 460W

voltage= 230V

Required

The amperage/ current of the fuse

Recall P= IV

I= P/V

I= 460/230

I=2 amps

Hence the current of the fuse is 2 amps

7 0

2 years ago

What is the velocity of a plane that traveled 3000 miles from new york to california in 5 hours

AnnZ [28]

The velocity is (3,000/5) = 600 miles per hour west .

5 0

3 years ago