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3 years ago

What are two pieces of information you must know about velocity on an object

1 answer:

3 0

Distance traveled(in a certain direction) and time
Velocity is all about direction traveled in comparison to speed which is just distance with out direction.

Hope this helps you

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A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

3 years ago

8–6. Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress in the cyli

vekshin1

Answer:

$P=2.66\ N$ is the maximum safe load.

Explanation:

Given:

diameter of the cylinder, $d= 90\ mm$

thickness of the cylinder wall, $t=2\ mm$

maximum bearable stress, $\sigma=3\ MPa$

Firstly we find that:

$\frac{t}{d} =\frac{1}{45}$

⇒ This is a thin cylinder.

We know axial stress and hoop stress is given by

$\sigma_a=\frac{P.d}{4t}$

$\sigma_h=\frac{P.d}{2t}$

⇒ Axial load will always be larger than the circumferential load under while other parameters are same.

So we find the load in case of axial stress:

$30=\frac{P\times 90}{4\times 2}$

$P=2.66\ N$ is the maximum safe load.

6 0

3 years ago

Read 2 more answers

A ball is thrown straight up in the air. When will its kinetic energy be the smallest before it is caught? at the start of its f

sergiy2304 [10]

Answer: At the top of its flight.

At the top of the flight,the height from the ground is maximum. Owing to this position the potential energy attained here is the maximum.

Also potential energy = mgh

Where m is the mass of the object.

g is the acceleration due to gravity

h is the height of object.

From this equation we can conclude that at the top,the maximum height would be attained and hence it would possess maximum potential energy.

Also at the top the object will possess zero velocity.

Kinetic energy = 1/2 (mv^2).

Where m is the mass of the object

v is the velocity of the object.

Hence at the top, since the velocity of the object is zero,the kinetic energy would be zero as explained in the below equation.

Kinetic energy= 1/2(m x 0)

Kinetic energy = 0.

Thus at the top of the flight,the ball possess only potential energy and minimum (0) kinetic energy.

4 0

4 years ago

A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 67 m/s2 for t1 = 39 s. The f

Igoryamba

Answer:

(a) $v_1= a_1t_1$

(b) $v_2 =a_1t_1+a_2t_2$

(c) 44133.5 m

Explanation:

Given:

$u$ = initial speed of the rocket in the first stage = 0 m/s
$v_1$ = final speed of the rocket in the first stage
$v_2$ = final speed of the rocket in the second stage
$t_1$ = time interval of the first stage
$t_2$ = time interval of the second stage
$s_1$ = distance traveled by the rocket in the first stage

$s_2$ = distance traveled by the rocket in the second stage
$s$ = distance traveled by the rocket in whole time interval

Part (a):

Since the rocket travels at constant acceleration.

$\therefore v_1 = u+a_1t_1\\\Rightarrow v_1 = a_1t_1$

Hence, the expression of the rocket's speed at time $t_1\ is\ v_1 = a_1t_1$ .

Part (b):

In this part also, the rocket moves with a constant acceleration motion.

$\therefore v_2 = v_1+a_2t_2\\\Rightarrow v_2 = a_1t_1+a_2t_2$

Hence, the expression of the rocket's speed in the time interval $t_2$ is $v_2 = a_1t_1+a_2t_2$ .

Part (c):

For the constant acceleration of rocket, let us first calculate the distance traveled by the rocket in both the time intervals.

$s_1 = u+\dfrac{1}{2}a_1t_1^2\\\Rightarrow s_1 = 0+\dfrac{1}{2}67\times(1)^2\\\Rightarrow s_1 =33.5\ m$

Similarly,

$s_2 = v_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = a_1t_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = 67\times1\times49+\dfrac{1}{2}\times 34\times(49)^2\\\Rightarrow s_2 =44100\ m\\\therefore s = s_1+s_2\\\Rightarrow s = 33.5\ m+44100\ m\\\Rightarrow s =44133.5\ m$

Hence, the rocket moves a total distance of 44133.5 m until the end of the second period of acceleration.

5 0

3 years ago

Read 2 more answers

A box of winter jackets of mass 109 kg is droped from a stationary helicopter at an altitude of 135

lesantik [10]

Answer:

tell me how and the explanation, I don't know it

3 0

3 years ago