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3 years ago

Why do solids have a definite shape?

2 answers:

5 0

Any matter that is a solid has a definite shape and a definite volume. The molecules in a solid are in fixed positions and are close together. Although the molecules can still vibrate, they cannot move from one part of the solid to another part. As a result, a solid does not easily change its shape or its volume

Rashid [163]3 years ago

4 0

"any matter that is a solid has a definite shape and a definite volume. the molecules in a solid are in fixed positions and are close together, although the molecules can still vibrate they cannot move from one part of the solid to another part. as a result, a solid does not easily change its shape or its volume."

I hope this helps ^-^

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In a simple series circuit, why does the bulb light up when you close the switch

photoshop1234 [79]

closing the switch completes the circuit

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4 years ago

What kind of weather system encourages a thunderstorm to develop

valentina_108 [34]

Low pressure systems

6 0

4 years ago

When radioactive tracers are used for medical scans, the tracers both decay in the body via radiation, as well as being excreted

irinina [24]

Answer:

Effective half-time of the tracer is 3.6 days

Explanation:

The formula for calculating the decay due to excretion for the first process is ;

$\frac{dN_1}{dt } = - \lambda _e N_o$

here ;

$N_o$ = initial number of tracers

Then to the second process ; we have :

$\frac{dN_2}{dt } = - \lambda _e N_o$

The total decay is as a result of the overall process occurring ; we have :

$\frac{dN_{total}}{dt } = \frac{dN_1}{dt} + \frac{dN_2}{dt}$ ------ (1)

here ;

$\frac{dN_{total}}{dt } = \lambda _{total} N_o$

Putting the values in (1);we have :

$- \lambda _{Total} N_o = - \lambda_e N_o + ( -\lambda r N_o})$

$\lambda _{Total} = \lambda_e + \lambda r$

As we also know that:

$\frac{1}{t_{1/2}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{9+6}{9*6}$

$\frac{1}{t_{1/2}_{effective}}}=\frac{15}{54}$

$t_{1/2}_{effective}} = \frac{54}{15}$

= 3.6 days

5 0

4 years ago

Read 2 more answers

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.40

leva [86]

Answer:

The work done required on the coin during the displacement is 21.75 w.

Explanation:

Given that,

A coin slides over a friction-less plane i.e friction force = 0.

The co-ordinate of the given point is (1.40 m, 7.20 m).

The position vector of the given point is represented by $1.40 \hat i+7.20 \hat j$ .

The displacement of the coin is

$\vec d=1.40 \hat i+7.20 \hat j$

The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.

Then x component of the force = 4.50 cos128°

The y component of the force = 4.50 sin128°

Then the position vector of the force is

$\vec F=(4.50 cos 128^\circ)\hat i+(4.50 sin 128^\circ)\hat j$

$=-2.77 \hat i+3.56 \hat j$

We know that,

work done is a scalar product of force and displacement.

$W=\vec F.\vec d$

$=(-2.77 \hat i+3.56 \hat j).(1.40 \hat i+7.20 \hat j)$

=(-2.77×1.40+ 3.56×7.20) w

=21.75 w

The work done required on the coin during the displacement is 21.75 w.

6 0

3 years ago

The area under the line and above the axes on a velocity vs. time graph represents what?

galben [10]

Area under the line and above the axis on a velocity - time graph represents the displacement of the object.

8 0

3 years ago