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Charra [1.4K]
3 years ago
8

The ability to distinguish between acceleration and velocity will be critical to your understanding of many other concepts in th

is course. Some of the most prevalent issues arise in interpreting the sign of both the velocity and acceleration of an object. I would recommend reading through the section "The Sign of the Acceleration" carefully. An object moves with a positive acceleration. Could the object be moving with increasing speed, decreasing speed or constant speed?
Physics
1 answer:
Sauron [17]3 years ago
7 0

Answer:

Object could only be moving with increasing speed.

Explanation:

Let us consider the general formula of acceleration:

a = (Vf - Vi)/t

Vf = Vi + at   -------- equation 1

where,

Vf = Final Velocity

Vi = Initial Velocity

a = acceleration

t = time

<u>FOR POSITIVE ACCELERATION:</u>

Vf = Vi + at

since, both acceleration and time are positive quantities. Hence, it means that the final velocity of the object shall be greater than the initial velocity of the object.

Vf > Vi

It clearly shows that if an object moves with positive acceleration. <u>It could only be moving with increasing speed.</u>

Solving the same equation for negative acceleration shows that the final velocity will be less than initial velocity and object will be moving with decreasing speed.

And for the constant velocity final and initial velocities are equal and thus, acceleration will be zero.

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1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche
Tatiana [17]

Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2

A^2=\dfrac{mv^2}{k}

Put the value into the relation

A^2=\dfrac{1.60\times13.0^2}{315}

A=\sqrt{0.858}

A=0.926\ m

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

a=A\omega^2

a=A\times\dfrac{k}{m}

Put the value into the formula

a=0.926\times\dfrac{315}{1.60}

a=182.31\ m/s^2

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

F=ma

Put the value into the formula

F= 1.60\times182.31

F=291.69\ N

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

7 0
3 years ago
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let’s find out by doing a
mixer [17]

Answer:

0.07142 m/s in the opposite direction of the bullet

Explanation:

m_1 = Mass of rifle = 70 kg

m_2 = Mass of bullet = 0.01 kg

v_1 = Velocity of rifle

v_2 = Velocity of bullet = 500 m/s

As the momentum of the system is conserved

m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.01\times 500}{70}\\\Rightarrow v_1=-0.07142\ m/s

The recoil velocity of the rifle is -0.07142 m/s

The negative sign shows the opposite direction of the rifle.

3 0
3 years ago
Assuming 100% efficient energy conversion, how much water stored behind a 50cm high hydroelectric dam would be required to charg
7nadin3 [17]

Answer:

12.245m3

Explanation:

The electric energy is created by The potential energy substended by the specific volume of water in dam.

Electric energy is calculated as

E= Q× V

E is Energy, Q is charge and V is Voltage

Note that this energy has been given and is 60Joules

From conservation of energy it means;

M× g×h = 60

Where M is the mass of water.

g is acceleration of free fall due to gravity which is 9.8m/S2

h is the height of water flow.

From change of subject of formula for M; we have:

M = 60/ g × h

= 60/ 9.8 × 0.5

= 12.245kg

Now how much water required means the volume of water;

Note density = mass/volume

Therefore volume = mass/density

=12.245/1= 12.245m3

Note the density of water is 1kg/m3

4 0
4 years ago
A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
maks197457 [2]

Answer:

5.6\times 10^{-8}\ Ohm.m

Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

8 0
4 years ago
10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.
Xelga [282]

Answer:

\lambda=1282\ nm

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

So,  

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

Given, n_i=5\ and\ n_f=3

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m

\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m

\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m

\lambda= 1.2828\times10^{-6}

1 m = 10⁻⁹ nm

\lambda=1282\ nm

6 0
3 years ago
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