What is the pressure exerted by a block of wood of mass 10 kg with an area of 0.1 m2

A person is diving in a lake in the depth of h = 5.5 m. The density of the water is rho = 1.0 x10^3 kg/m^3. The pressure of the

Cloud [144]

Answer:a) P = Po + rho×h×g

b) P = 5.4 × 10^9 pa

c) F = P/A = (Po + rho×h×g)/A

d) 1.174×10^11N

Explanation: Using the formula

P = Po + rho×h×g

P = 1.0 x 10^5 + 1000 × 5.5 × 9.81

P = 5.4 × 10^9pa

The magnitude of the force exerted by water on the top of the person's head F at the depth h in terms of P

F = P/A = (Po + rho×h×g)/A

Using the above formula

Where A = 0.046m^2

F = P/ A = 5.4×10^9/0.046

F = 1.174×10^11N

3 0

2 years ago

The scientist who found a way to measure the distance between the sun and Venus which

Yakvenalex [24]

Answer:

Astronomer Edmond Halley

Explanation:

The astronomical unit using the transit of venus

The underlying principle behind Halley's method is called parallax

8 0

2 years ago

QUESTION 10

Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

$T_{1}sin(\theta)-T_{2}sin(\theta)=0$ (1)

Where:

T(1) is the tension due to the rope 1
T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0

2 years ago

The impact that an earthquake can have on an area?

tatyana61 [14]

Answer:

The primary effects of earthquakes are ground shaking, ground rupture, landslides, tsunamis, and liquefaction. Fires are probably the single most important secondary effect of earthquakes.

Explanation:

7 0

2 years ago

Read 2 more answers

A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

6 0

2 years ago