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OleMash [197]
2 years ago
7

Which of the following is closest to 2cm?

Physics
1 answer:
Sloan [31]2 years ago
4 0
C because the the smallest thing then the other ones because it never said what kind of size of it
You might be interested in
Calculate the hydrostatic difference in blood pressure between the brain and the foot in a person of height 1.93 m. The density
Slav-nsk [51]

Answer:

Explanation:

Given: Density of blood = 1.03 × 10³ Kg/m³, Height =  1.93 m g = 9.8 m/s²

pressure at the brain is equal to atmospheric pressure. = Hydro-static

pressure(ρ₀)

∴ pressure of the foot = pressure of the brain(ρ₀) + ( density of blood × acceleration due to gravity × height)(ρgh)

Hydro-static pressure = pressure at the feet- pressure at the brain(ρ₀)

Hydro-static pressure (Δp) = (ρgh + ρ₀) - ρ₀ = ρgh

Hydro-static pressure = 1.03 × 10³ × 9.8 × 1.93 = 1.948 × 10⁴ Pa

∴  Hydro-static pressure ≈ 1.95 × 10⁴ Pa

3 0
2 years ago
A spaceship is headed toward Alpha Centauri at 0.999c. According to us, the distance to Alpha Centauri is about 4 light-years. H
aniked [119]

Answer:

According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>

<em></em>

Explanation:

For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:

v = d / t

where

  • v is the speed of the spaceship
  • d is the distance
  • t is the time

Therefore,

d = v × t

d = (0.999 c)(4 light-years)

d = 3.996  light-years

This distance is<em> very slightly less than 4 light-years. </em>

4 0
3 years ago
I got part c right but idk why the other parts are wrong HELP!
dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

J=\int F\Delta t

where

F is the magnitude of the force exerted on the object

\Delta t is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

B=0.14-0 = 0.14s\\b=8-5=3s

and whose height is

h=900 N

Therefore, the area (and the impulse) is

J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

J=F_{avg} \Delta t

where

F_{avg} is the average force exerted during the whole time \Delta t

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

\Delta t = 0.14 s is the time interval

Solving for the average force, we find

\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

J=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0
3 years ago
True or False. The pressure inside a piston is the same in all directions.
Dominik [7]

Answer:

True

Explanation:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at any point in the fluid is equal in all directions.

4 0
2 years ago
A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t
Talja [164]

Answer:

P = 133.13 Watt

Explanation:

Initial angular speed of the ferris wheel is given as

\omega_i = 2\pi f

\omega_i = 2\pi(8.5/3600)

\omega_i = 0.015 rad/s

final angular speed after friction is given as

\omega_f = 2\pi f

\omega_f = 2\pi(7.5/3600)

\omega_f = 0.013 rad/s

now angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{0.015 - 0.013}{15}

\alpha = 1.27 \times 10^{-4} rad/s^2

now torque due to friction on the wheel is given as

\tau = I \alpha

\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})

\tau = 8875.3 N m

Now the power required to rotate it with initial given speed is

P = \tau \omega

P = 8875.3 \times 0.015

P = 133.13 Watt

8 0
2 years ago
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