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3 years ago

Define scalar and vector

1 answer:

7 0

The quantity is either a vector or a scalar. These two categories can be distinguished from one another by their distinct definitions: Scalars are quantities that are fully described by a magnitude (or numerical value) alone. Vectors are quantities that are fully described by both a magnitude and a direction.

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The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a

Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

with m the mass , a the acceleration and $\sum\overrightarrow{F}$ the net force forces that is:

$(F-f)$ (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

$(F-f)=ma$

solving for a:

$a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}$

Now let's use the Galileo’s kinematic equation

$Vf^{2}=Vo^{2}+2a\varDelta x$ (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and $\varDelta x$ the time took to achieve that velocity, solving (3) for $\varDelta x$ :

$\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}$

$t=68.94 m$

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3 years ago

A 3 kg object is attached to a 1000 N/m spring. The spring is compressed 0.10 m and then the spring launches the object horizont

Viktor [21]

Answer:

1.826m/s

Explanation:

E=1/2*k*(∆L)^2

E=1/2*mV^2

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3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

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3 years ago

Energy conversion is the change of energy from ____ to another.

Alexeev081 [22]

from one energy form to one

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3 years ago

Compare the distance-time graph for a fast-moving object with the distance time graph of a slower moving object.

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The distance vs time graph of a faster moving object would have a greater slope than that of a slower moving one. The object would move a greater distance per unit time

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3 years ago