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2 years ago

Why don't you feel the pressure exerted by the atmosphere?

Physics

2 answers:

mariarad [96]2 years ago

8 0

Answer:

the air within our bodies (in our lungs and stomachs, for example) is exerting the same pressure outwards so there's no pressure difference and no need for us to exert any effort.

Explanation:

Galina-37 [17]2 years ago

6 0

Because in our body blood also exerts pressure that's why we don't feel atmospheric pressure.

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If a football player hits the ball with a force of 50 N, determine the reaction force.

mel-nik [20]

Answer: 60

Explanation: if u hit it it will have impact and the impact added 10n

4 0

2 years ago

An object moving at 15 m/s slows uniformly at a rate of 2.0 m/s each second for 5.0 s. What is its final speed?

NeX [460]

Hey there!

We are given ,

Acceleration, a = -2m/s^2

Initial velocity , u = 15m/s

Time , t = 5 seconds

We know that ,
V=u+at

Now , final speed ,

V = 15+(-2)(5)

V = 15-10

V = 5 m/s -> final speed

Hope this helps you dear :)
Have a good day <3

8 0

1 year ago

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

USPshnik [31]

Answer:

a) see attached, a = g sin θ

c) v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

Wₓ = m a

W sin θ = m a

a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

θ' = 9/2 θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

Em₀ = mg h = mg L (1-cos tea)

Lowest point

Emf = K = ½ m v²

Em₀ = Emf

g L (1-cos θ) = v² / 2

v = √(2gL (1-cos θ))

4 0

2 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$