All categories

3 years ago

What should you do if...

Physics

1 answer:

Hunter-Best [27]3 years ago

6 0

Answer:

3. you need to ask your available lab instructors what to do.

4. You immediately have to drop down your cloth and roll it to extinguish the fire or move to the emergency shower if available

5. You have too keep calm and report to the lab instructor but do no shout.

6. Move immediately to the eye rinse basin if available and wash your eyes gently and thoroughly

You might be interested in

The electromagnetic wave that delivers a cellular phone callto

spin [16.1K]

Answer:

$E=2.41\cdot 10^{-5} J$

Explanation:

The intensity of an electromagnetic wave can be expressed in terms of the magnetic field using the next relationship:

$I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}}$ (1)

c is the speed of light (3*10⁸ m/s)
μ₀ is the permeability of free space (in vacuum ) (1.26*10⁻⁶ N/A²)
B₀ is the magnetic field

$I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}$

$I_{average}=2.68\cdot 10^{-6} W/m^{2}$

Now, let's define the relationship between power (P) and average intensity (I).

$I_{average}=\frac{P}{A}$

P is the power
A is the area crossed

So we can calculate the power.

$P=I_{average}\cdot A=2.68\cdot 10^{-6}\cdot 0.20=5.37\cdot 10^{-7} W$

Finally, energy is the product of P times time, so:

$E=P\cdot t=5.37\cdot 10^{-7} \cdot 45=2.41\cdot 10^{-5} J$

I hope it helps you!

5 0

2 years ago

Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached

RUDIKE [14]

Answer:

Wmax = 63.65 ≈ 64 lb

Explanation:

6 0

3 years ago

If a spacecraft is pushed by a thruster, what will happen

ElenaW [278]

D.

This is an accordance with Newton's First Law of Motion, stating that objects in motion remain in motion and objects at rest remain at rest unless acted upon by an external force.

8 0

3 years ago

A block of mass m moving with a velocity v collides with a mass 2m at rest. Consider the collision is elastic, we have to find t

lions [1.4K]

Answer:

$vf_{1} =\frac{1}{3} v$ :Final block velocity (toward the left)

$vf_{2} =\frac{2}{3} v$ :Final mass velocity (to the right)

Explanation:

To solve this problem we apply the theory of shocks:

In an elastic shock the kinetic energy and the amount of linear movement or momentum are conserved.

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

Principle of conservation of the momentum:

m1vi1+m2vi2=m1vf1+m2vf2 Equation 1

Formula to calculate the coefficient of elastic restitution (e):

$e=\frac{vf_{2}-vf_{1} }{vi_{1}-vi_{2} }$ Equation 2

m1: Block mass

m2: mass of the body that collides with the block

Vi1,vf1: initial, final velocity of the block

Vi2,vf2: initial, final velocity of the body that collides with the block

Of the problem data we know that:

m1=m , m2 = 2m, vi1=v and vi2=0, then, we replace this information in equation (1) :

mv+0=mvf1+2mvf2 we divide by m:

v=vf1+2vf2 Equation (3)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1,then , we we replace this information in equation (2)

$1=\frac{vf_{2}-vf_{1} }{v-0}$

v=vf2-vf1 Equation (4)

vf2=v+vf1 Equation (5)

We replace the equation 5 in the equation (3)

v=vf1+2(v+vf1)

v=vf1+2v+2vf1

-3vf1=v

$vf_{1} = -\frac{1}{3} v$

We replace Vf1=(-1/3)v in the equation (5):

$vf_{2} =v-\frac{1}{3} v$

$vf_{2} =\frac{2}{3} v$

Answer;

$vf_{1} =\frac{1}{3} v$ :Final block velocity (toward the left)

$vf_{2} =\frac{2}{3} v$ :Final mass velocity (to the right)

5 0

2 years ago

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, while a

xxTIMURxx [149]

Answers:

a) $F_{g}=735 N$ and $n=732.47 N$ , hence $F_{g} > n$

b) $n_{poles}=735 N$ $n_{equator}=732.47 N$

Explanation:

a) At the equator, both the centripetal force $F_{c}$ and the gravitational force $F_{g}$ (also called true weight) are directed "downward", while the normal force $n_{equator}$ (also called apparent weight) is directed "upward". Therefore we have the following equation: