All categories

3 years ago

Acceleration is equal to the initial velocity minus the final velocity, then divided by time.true or false

Physics

1 answer:

tatiyna3 years ago

5 0

Almost true but not quite.

That would give you the negative of the actual acceleration.

It should be the other way around:

(final v) minus (initial v), then divide by time.

You might be interested in

What is the kinetic energy of a 4 kg toy car moving at 5 m/s?

Mamont248 [21]

Answer:

50 J

Explanation:

4 0

3 years ago

A 1200 kg aircraft going 30 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and

Leno4ka [110]

Answer:

83.055 m

Explanation:

According to the given scenario, the calculation of skid distance is shown below:-

$S = \frac{1}{2} \times (u + v) \times t$

Where

u = 11.3

v = 0

t = 14.7

Now placing these values to the above formula,

So,

$S = \frac{1}{2} \times (11.3 + 0) \times 14.7$

= 83.055 m

Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question

4 0

3 years ago

The force of gravity is less between two objects that

Aloiza [94]

The force of gravity is less between to objects when the product
of their masses is smaller, or when they are farther apart.

6 0

3 years ago

When light hits a mirror it bounces off of it. This is why you can see yourself in a mirror. This property of light is A) absorp

son4ous [18]

I think it's B reflection

8 0

3 years ago

Read 2 more answers

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:

<u>Given:</u>

$u$ = 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