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3 years ago

13

Give two examples of common force fields and name the sources of these fields.

1 answer:

algol133 years ago

4 0

Electric force from electomagnetic force and force of gravity from gravitational force

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A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The

Ber [7]

Answer:

1. $v=14.2259\ m.s^{-1}$

2. $F_T=25.8924\ N$

3. $\lambda=29.6373\ m$

Explanation:

Given:

mass of slinky, $m=0.87\ kg$

length of slinky, $L=6.8\ m$

amplitude of wave pulse, $A=0.23\ m$

time taken by the wave pulse to travel down the length, $t=0.478\ s$

frequency of wave pulse, $f=0.48\ Hz=0.48\ s^{-1}$

1.

$\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel$

$v=\frac{6.8}{0.478}$

$v=14.2259\ m.s^{-1}$

2.

<em>Now, we find the linear mass density of the slinky.</em>

$\mu=\frac{m}{L}$

$\mu=\frac{0.87}{6.8}\ kg.m^{-1}$

We have the relation involving the tension force as:

$v=\sqrt{\frac{F_T}{\mu} }$

$14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }$

$202.3774=F_T\times \frac{6.8}{0.87}$

$F_T=25.8924\ N$

3.

We have the relation for wavelength as:

$\lambda=\frac{v}{f}$

$\lambda=\frac{14.2259}{0.48}$

$\lambda=29.6373\ m$

8 0

2 years ago

According to the definition of mechanical work, pushing on a rock accomplishes no work unless there is

LekaFEV [45]

There must be movement in the same direction as the force put on the object. Hope this helps!

4 0

3 years ago

Read 2 more answers

-!!! URGENT !!!-

seraphim [82]

Try this option, the answers are marked with colour.

8 0

3 years ago

Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as show

Grace [21]

Answer:

pretty sure its motor 2, with body 1

Explanation:

4 0

3 years ago

Read 2 more answers

Suppose a clay model of a koala bear has a mass of 0.205 kg and slides on ice at a speed of 0.720 m/s. It runs into another clay

Westkost [7]

Answer:

Final velocity, v = 0.28 m/s

Explanation:

Given that,

Mass of the model, $m_1=0.205\ kg$

Speed of the model, $u_1=0.72\ m/s$

Mass of another model, $m_2=0.32\ kg$

Initial speed of another model, $u_2=0$

To find,

Final velocity

Solution,

Let V is the final velocity. As both being soft clay, they naturally stick together. It is a case of inelastic collision. Using the conservation of linear momentum to find it as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{0.205\times 0.72+0.32\times 0}{(0.205+0.32)}$

V = 0.28 m/s

So, their final velocity is 0.28 m/s. Hence, this is the required solution.

6 0

2 years ago