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3 years ago

14

You take the mass of the object 400 grams then what do you predict the force needed to move the object at a constant acceleratio

n of 3/s2?

1 answer:

Oksanka [162]3 years ago

6 0

Answer:

Below

Explanation:

To find the force needed to move an object, you can use this formula :

force = (mass)(acceleration)

Plugging our values in...

force = (400g)(3 m/s^2)

= 1,200 Newtons

We can see how this works for the previous answers :

force = (100g)(3 m/s^2)

= 300 Newtons and so on....

Hope this helps! Best of luck <3

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Which statement best describes the equation CaCO3 + 2HCl → CaCl2 + CO2 + H2O? CaCO3 is a reactant; it is present before the reac

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Answer:

CaCl2 is a reactant

Explanation:

Calcium carbonate (CaCO3) is a reagent, it is found on the left side, what is found on the right side are the products.

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In damped harmonic oscillation, the amplitude of oscillation becomes one third after 2 second. If A0 is initial amplitude of osc

Harman [31]

Answer:

$A=\frac{A_0}{\sqrt 3}$

Explanation:

Initial amplitude= $A_0$

We are given that

Amplitude after 2 s=A= $\frac{1}{3}A_0$

We have to find the amplitude after 1 s.

We know that amplitude at any time t

$A=A_0e^{-\alpha t}$

Using the formula

$\frac{A_0}{3}=A_0e^{-2\alpha}$

$\frac{1}{3}=e^{-2\alpha}$

$3=e^{2\alpha}$

$ln 3=2\alpha$

$\alpha =\frac{ln 3}{2}=ln\sqrt 3$

$e^{\alpha}=\sqrt 3}$

When t=1 s

$A=A_0e^{-\alpha}=\frac{A_0}{\sqrt 3}$

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4 years ago

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$

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3 years ago

After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.40 rad/s and it rotated 12.3 re

zaharov [31]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of angular motion.

By definition, acceleration can be expressed as the change in angular velocity squared over a given period of distance traveled.

$\alpha = \frac{\omega^2}{2\theta}$

where,

$\omega =$ Angular velocity

$\theta =$ Angular displacement.

In turn, as a function of time, we can represent it as,

$\alpha = \frac{\omega}{t}$

For our case we have to,

$\omega = 5.4rad/s$

$\theta = 12.3rev = 12.3rev(\frac{2\pi rad}{1rev})=24.6\pi rad$

PART A) In the case of angular acceleration we have to,

$\alpha = \frac{\omega^2}{2\theta}$

$\alpha = \frac{(5.4)^2}{2*24.6\pi}$

$\alpha = 0.1886rad/s^2$

PART B) Through the definition of angular acceleration as a function of time we can calculate it,

$\alpha = \frac{\omega}{t}$

$t = \frac{\omega}{\alpha}$

$t = \frac{5.4}{0.1886}$

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4 years ago

An elastic conducting material is stretched into a circular loop of 14.7 cm radius. It is placed with its plane perpendicular to

Hunter-Best [27]

Answer: 0.66 V

Explanation:

Given

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Instantaneous rare = 74.5 cm/s = 0.745 m/s

radius, r = 14.7 cm = 0.147 m

We will use the formula

emf = dΦ/dt

emf = d(BA)/dt

emf = d(Bπr²)/dt

if B is constant, then we can say

emf = Bπ d(r²)/dt on differentiating, we have,

emf = Bπ (2r dr/dt)

emf = 2πrB dr/dt substituting each values, we have

emf = 2 * 3.142 * 0.147 * 0.963 * 0.745

emf = 0.66 V

Therefore, the induced emf in the loop at that instant is 0.66 V

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3 years ago