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2 years ago

Which statements are part of the safety protocol for this lab experiment? Check all that apply.

Physics

2 answers:

tekilochka [14]2 years ago

8 0

Answer:

Always wear safety goggles when performing an experiment.

Use caution when dropping objects and/or launching them into the air.

Use only the materials that your teacher provides or approves.

Check the soda bottle and its cap for cracks and chips prior to use.

Report all accidents—no matter how big or small—to your teacher.

Mazyrski [523]2 years ago

5 0

Answer:

1. Always wear a lab coat and safety goggles when performing an experiment

2. Use the right gear, such as tongs and thermal mitts, to handle hot objects

3. Check glassware, such as beakers and flasks, for cracks and chips prior to use.

Explanation:

It is in the safety section of the student guide for edge.

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Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

2 years ago

PLEASE HELP ME ON THIS ONE ANYONE

andreev551 [17]

Answer:

A & B

Explanation:

A & B Would be the right answer since Morse code cannot be represented through the height of the fire.

6 0

1 year ago

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

3 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

2 years ago

Coach Hogue is riding his motorcycle in a circle on wet pavement. Suddenly the bike slides out from under him. What failed to pr

Arisa [49]

Answer:

The correct option is;

Force of Friction

Explanation:

As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path

The motion equation is

v = ωr and we have the centripetal acceleration given by

α = ω²r and therefore centripetal force is then

m×α = m × ω²r = m × v²/r

The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.

5 0

3 years ago