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3 years ago

Which is an indication that a man might be choking

Physics

2 answers:

Maksim231197 [3]3 years ago

5 0

Answer:

if they are struggling to breathe, attempting to cough (like coughing something up) or asking for help whilst holding their stomach, neck

nalin [4]3 years ago

4 0

Answer:

If they put their hands near their neck, if they're not breathing properly, if they aren't coughing.

Explanation:

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A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$

$F_{t}-m.g=m.a$

$18.8-9.8m=5.22m$

$15.02m=18.8$

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: $F_{g}=m.g$ .

The weight for the hanging block is

$F_{g}=1.25*9.8$

$F_{g}=$ 12.25 N

Comparing tension and weight:

$\frac{12.25}{18.8}$ ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

4 0

3 years ago

A. Obtain the following: microwave, ruler, something meltable (e.g. candy bar, marshmallows) B. A microwave works by setting up

Makovka662 [10]

Answer:

E = 124.7 N / C

Explanation:

Let's analyze the exercise: the microwave creates an electromagnetic wave of frequency F = 2.45 GHz, this wave is introduced into the microwave cavity and is reflected on the metal walls, which is why one or more standing waves are formed.

The electric field of the standing wave is

I = E²

E =√I

where I is the intensity of the radiation.

What is it

I = P / A

where P is the effective emission power, almost all the power of the microwave and A is the area of the cavity, in the most used microwaves

P = 700 W and the area is A = 25 x 18 cm² = 0.045 m²

I = 700 / 0.045

I = 15555.56 W/m²

let's calculate the electric field

E = √15555.56

E = 124.7 N / C

7 0

3 years ago

Example of kinetic energey

liubo4ka [24]

Answer:

rolling ball down a hill

Explanation:

A rolling ball has kinetic energy

5 0

3 years ago

Read 2 more answers

How high was a brick dropped from if if falls in 2.5 seconds?

jenyasd209 [6]

Using the kinematic equation d = V_0 * t + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
d = 30.7m

7 0

3 years ago

A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha

USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

$\Phi_E=\frac{Q}{\epsilon_o}$

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

$\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}$

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

$Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C$

Finally, you obtain for E:

$E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}$

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0

3 years ago