Answer:
c. A number and a unit
Explanation:
A scalar is a real number. We often use the term scalar in the context of vectors or matrices, to stress that a variable such as a is just a real number and not a vector or matrix.
Answer:
daughters cells
Explanation:
when people refer to “cell division,” they mean mitosis, the process of making new body cells. Meiosis is the type of cell division that creates egg and sperm cells. Mitosis is a type of cell division in which one cell (the mother) divides to produce two new cells (the daughters) that are genetically identical to itself.
Answer:
P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,
Explanation:
This problem of fluid mechanics let's start with the continuity equation to find the speed of water output
Q = A v
v = Q / A
The area of a circle is
A = π r² = π d² / 4
Let's look at the speeds at each point
v₁ = Q / A₁ = Q 4 /π d₁²
v₁ = 10 4 /π 0.5²
v₁ = 50.93 m / s
v₂ = Q / A₂
v₂ = 10 4 /π 0.25²
v₂ = 203.72 m / s
Now we can use Bernoulli's equation in the colon
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear
P₁ = P2 + ½ rho (v₂² - v₁²)
P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)
P₁ = 1.013 10⁵ + 2.205 10⁷
P₁ = 2.215 10⁷ Pa
la definicion de presion es
P₁ = F₁/A₁
F₁ = P₁ A₁
F₁ = 2.215 10⁷ pi d₁²/4
F₁ = 2.215 10⁷ pi 0.5²/4
F₁ = 4.3 106 N
Answer:
a. ![t_1=12.5\ s](https://tex.z-dn.net/?f=t_1%3D12.5%5C%20s)
b.
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,
![u_1=25\ m.s^{-1}](https://tex.z-dn.net/?f=u_1%3D25%5C%20m.s%5E%7B-1%7D)
- speed of lagging car,
![u_{2}=35\ m.s^{-1}](https://tex.z-dn.net/?f=u_%7B2%7D%3D35%5C%20m.s%5E%7B-1%7D)
- distance between the cars,
![\Delta s=45\ m](https://tex.z-dn.net/?f=%5CDelta%20s%3D45%5C%20m)
- deceleration of the leading car after braking,
![a_1=-2\ m.s^{-2}](https://tex.z-dn.net/?f=a_1%3D-2%5C%20m.s%5E%7B-2%7D)
a.
Time taken by the car to stop:
![v_1=u_1+a_1.t_1](https://tex.z-dn.net/?f=v_1%3Du_1%2Ba_1.t_1)
where:
, final velocity after braking
time taken
![0=25-2\times t_1](https://tex.z-dn.net/?f=0%3D25-2%5Ctimes%20t_1)
![t_1=12.5\ s](https://tex.z-dn.net/?f=t_1%3D12.5%5C%20s)
b.
using the eq. of motion for the given condition:
![v_2^2=u_2^2+2.a_2.\Delta s](https://tex.z-dn.net/?f=v_2%5E2%3Du_2%5E2%2B2.a_2.%5CDelta%20s)
where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking
![0^2=35^2+2\times a_2\times 45](https://tex.z-dn.net/?f=0%5E2%3D35%5E2%2B2%5Ctimes%20a_2%5Ctimes%2045)
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:
![v_2=u_2+a_2.t_2](https://tex.z-dn.net/?f=v_2%3Du_2%2Ba_2.t_2)
![0=35-13.61\times t_2](https://tex.z-dn.net/?f=0%3D35-13.61%5Ctimes%20t_2)
is the time taken to stop after braking
"<span>a layer in the earth's stratosphere at an altitude of about 6.2
miles (10 km) containing a high concentration of ozone, which absorbs
most of the ultraviolet radiation reaching the earth from the sun."
Hope this helps!
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