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3 years ago

13

A block of ice is floating in a liquid of specific gravity 1.2 if the ice melts completely will the level of water in the beaker

changes? And why?

1 answer:

yulyashka [42]3 years ago

4 0

Answer:

$Yes$

Explanation:

The way that it will change is by level of liquid in the whatever you are using will rise. This is because of the density of water created by the melting process of ice has a lower amount than the density of liquid in the beaker.

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You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pol

kozerog [31]

Answer:

$F=133N$

Explanation:

From the question we are told that:

Length $l=3.0m$

Mass $m=24kg$

Distance from Tip $d=35cm$

Generally, the equation for Torque Balance is mathematically given by

$mg(l/2)=F(l-d)$

$2*9.81(3/2)=F(3-35*10^-2)$

Therefore

$F=133N$

8 0

3 years ago

Compare and contrast offensive and defensive roles in team sports.

strojnjashka [21]

In the offensive role, the players try to get a goal.
In the defensive roll, The players try to protect the goal
Hoped this helped a little :)

4 0

3 years ago

Read 2 more answers

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:
</span>
$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span> $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

6 0

3 years ago

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0

3 years ago

ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr

Free_Kalibri [48]

Answer:

$\beta = 41.68°$

Explanation:

according to snell's law

$\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }$

refractive index of water n_w is 1.33

refractive index of glass n_g is 1.5

$sin\alpha = \frac{n_w}{n_g}* sin30$

$sin\alpha = 0.443$

now applying snell's law between air and glass, so we have

$\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}$

$sin\beta = \frac{n_g}{n_a} sin\alpha$

$\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]$

we know that $sin\alpha = 0.443$

$\beta = 41.68°$

7 0

3 years ago