All categories

3 years ago

Why is acceleration measured in meters / seconds squared? plz use your own words

2 answers:

8 0

Acceleration is velocity over time, and velocity is displacement over time. (This explains why time is squared.) As for the units, physicists (as well as pretty much any other kind of scientist) rely on SI units, which are the units generally accepted by scientists world-wide. Meters are the SI unit for length, and seconds are the SI unit for time.

oksano4ka [1.4K]3 years ago

6 0

This is because to find acceleration, you must divide velocity by time. Velocity is measured in meters / second, and time is measured in seconds. So, when meters / second is divided by seconds, the units become meters / second squared.

Hope this helps

Tobey

You might be interested in

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0

3 years ago

A cat with a mass of 4.50 kilograms sits on a ledge 0.800 meters above the ground. If it jumps to the ground, how much kinetic e

Slav-nsk [51]

Answer:

The cat will have <span>36J</span> of kinetic energy.

5 0

3 years ago

Read 2 more answers

Pete is driving down 7th Street. He drives 300 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe

Korolek [52]

Answer:

16.67m/s

Explanation:

Given parameters:

Distance Pete drove = 300m

Time taken = 18s

Unknown:

Speed = ?

Solution:

Speed is the distance traveled per unit of time.

It is mathematically expressed as;

Speed = $\frac{distance}{time}$

Insert the parameters and solve;

Speed = $\frac{300}{18}$ = 16.67m/s

3 0

3 years ago

Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 4R. System A consists of two of

Harman [31]

Answer:

4 smaller disks

Explanation:

We are given;

Mass of smaller and larger disks = M

Radius of smaller disk = R

Radius of larger disk = 4R

Formula for moment of inertia about cylinder axis is:

I = ½MR²

Thus;

For small disk, I_small = ½MR²

For large disk, I_large = ½M(2R)² = 2MR²

We are told that moment of inertia of System A consists of two of the larger disks. Thus;

I_A = 2 × I_large = 2 × 2MR²

I_A = 4MR²

We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;

I_B = I_large + n(I_small)

Where n is the number of smaller disks.

I_B = 2MR² + n(½MR²)

I_B = MR²(2 + n/2)

We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;

I_A = I_B

So;

4MR² = MR²(2 + n/2)

MR² will cancel out to give;

4 = 2 + n/2

Multiply through by 2 to give;

8 = 4 + n

n = 8 - 4

n = 4

5 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago