All categories

4 years ago

PLEASE HELP WILL GIVE BRAINLIEST!!!!!!!!

2 answers:

6 0

use a = (v1-v2)/t so u would put 14 (initial velocity or v 1) - 6 (the v2) and divide all of that by 4 seconds to get 2m/s/s as your acceleration

Naddik [55]4 years ago

6 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (starting speed)

Change in speed = (14 m/s) - (6 m/s) = 8 m/s

Time for the change = 4 seconds

Acceleration = (8 m/s) / (4 sec)

Acceleration = (8/4) (m/s·s)

Acceleration = 2 m/s²

You might be interested in

True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.

bazaltina [42]

The answer is true.

5 0

3 years ago

Read 2 more answers

A drag racer starts from rest and accelerates at 7.4 m/s2. How far will he travel in 2.0 seconds?

3241004551 [841]

Using the kinematic equation below we can determine the distance traveled if t=2, a=7.4m/s^2. First we must determine the final velocity:

$v_{final}=v_{initial}+\frac{1}{2}at\\\\v_{final}=0+(7.4m/s^2)(2s)=34.8m/s$

Now we will determine the distance traveled:

$v_{final}^2=v_{initial}^2+2a \Delta x\\\\\Delta x = \frac{v_{final}^2}{2a} =\frac{(34.8)^2}{(2)(7.4)}=81.83 m$

Therefore, the drag racer traveled 81.83 meters in 2 seconds.

4 0

3 years ago

Select three possible applications of a capacitor. Select all that apply.

Wittaler [7]

Answer:

I'm pretty sure it's all of them i'm not completely sure though hope it helps anyways! :)

7 0

4 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

mina [271]

Answer:

(a) v = 463.97 m/s, $a_{c} = 0.0337\ m/s^{2}$

(b) v' = 196.02 m/s, $a'_{c} = 0.0143\ m/s^{2}$

Solution:

As per the question:

Time period of the rotation of earth, T = 24 h = $24\times 3600 = 86400\ s$

Radius of the earth, R = $6.38\times 10^{6}\ m$

Angle, $\theta = 65.0^{\circ}$

Now,

Angular velocity is given by:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{86400} = 7.27\times 10^{- 5} \rad$

(a) To calculate the speed and the centripetal acceleration at the equator:

Linear velocity or the speed, v = $R\omega$

$v = 6.38\times 10^{6}\times 7.27\times 10^{- 5} = 463.967\ m/s$

Centripetal Acceleration, $a_{c} = \frac{v^{2}}{R}$

$a_{c} = \frac{463.967^{2}}{6.38\times 10^{6}} = 0.0337\ m/s^{2}$

(b) To calculate the speed and acceleration at an altitude of $65.0^{\circ}$ N:

The horizontal component of the radius, R' = $Rcos\theta$

$R' = 6.38\times 10^{6}cos65.0^{\circ} = 2.69\times 10^{6}\ m$

Now,

For the speed, v' = $R'\omega = 2.69\times 10^{6}\times 7.27\times 10^{- 5} = 196.02\ m/s$

For the centripetal acceleration,

$a'_{c} = \frac{v'^{2}}{R'}$

$a'_{c} = \frac{196.02^{2}}{2.69\times 10^{6}} = 0.01428\ m/s^{2}$

8 0

4 years ago

In the formula Work=Force X Distance, what happens to the amount of work done if either Force or Distance is increased?

AfilCa [17]

Explanation:

If force or distance is increased, then amount of workdone will also increase.

6 0

3 years ago