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2 years ago

Which describes the position on a horizontal number line

Physics

1 answer:

Savatey [412]2 years ago

3 0

Answer:

It is formed by a horizontal number line, called the x-axis, and a vertical number line, called the y-axis.

Explanation:

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A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an

Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0

3 years ago

Read 2 more answers

What includes a distance and a direction? A. Displacement B. Velocity C. Speed D. Acceleration

iren [92.7K]

hi <3

the correct option would be A. displacement. displacement is distance in a direction

hope this helps :)

7 0

2 years ago

Read 2 more answers

20 pts !!!!

garri49 [273]

Am not really sure but what i see is D

4 0

2 years ago

Verify that the SI unit of impulse is the same as the SI unit of momentum.

lys-0071 [83]

Maybe this will help you out:

Momentum is calculate by the formula:

$P = mv$

Where:

P = momentum

m = mass

v = velocity

The SI unit:

$mass = kg\\ velocity = \dfrac{m}{s}$

So the unit of momentum would be:

$kg.\dfrac{m}{s}$

Impulse is defined as the change in momentum or how much force changes momentum. It can be calculate with the formula:

I = FΔt

where:

I = impulse

F = Force

Δt = change in time

The SI unit:

F = Newtons (N) or $kg.\dfrac{m}{s^{2} }$

t = Seconds (s)

So the unit of impulse would be derived this way:

I = FΔt

I = $kg.\dfrac{m}{s^{2} }$ x $s$

$\dfrac{kg.m.s}{s^{2}} = \dfrac{kg.m.s}{s.s}$

You can then cancel out one s each from the numerator and denominator and you'll be left with:

$kg.\dfrac{m}{s}$

So then:

Momentum: Impulse

$kg.\dfrac{m}{s}$ $kg.\dfrac{m}{s}$

4 0

3 years ago

a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d

USPshnik [31]

Answer:

The time is $t = 2.595 \ s$

The speed is $v = 25.43 \ m/s$

Explanation:

From the question we are told that

The height of the cliff is $h = 33 \ m$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} gt^2$

before the jump the persons initial velocity is u = 0 m/s

$33 = 0 * t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.595 \ s$

Generally from kinematic equation

$v= u + gt$

=> $v= 0 + 9.8 * 2.595$

=> $v = 25.43 \ m/s$

3 0

3 years ago