HELPP!! what is not true about mitosis??

vovangra [49]

Answer:

X-rays are commonly produced in X-ray tubes by accelerating electrons through a potential difference (a voltage drop) and directing them onto a target material

5 0

2 years ago

An airplanes lands with a velocity of +75 m/s and comes to a stop in 20 seconds. What is the acceleration of the airplane while

erma4kov [3.2K]

Answer:

The acceleration of the car, a = -3.75 m/s²

Explanation:

Given data,

The initial velocity of the airplane, u = 75 m/s

The final velocity of the plane, v = 0 m/s

The time period of motion, t = 20 s

Using the I equations of motion

v = u + at

a = (v - u) / t

= (0 - 75) / 20

= -3.75 m/s²

The negative sign indicates that the plane is decelerating

Hence, the acceleration of the car, a = -3.75 m/s²

7 0

3 years ago

What does it mean when a wave’s amplitude increases?

sashaice [31]

Answer:

the wave is carrying more energy

Explanation:

trust me broski

7 0

2 years ago

As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce

WINSTONCH [101]

Answer:

a) x₁ = 290.50 feet , x₂ = 169.74 feet , b) v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis fr = m a

Y Axis N-W = 0

N = W = mg

The force of friction has the expression

fr = μ N

We replace

μ mg = ma

a = μ g

g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ a (feet / s2)

0.599 19.168

0.536 17,152

0.480 15.360

0.350 11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

v² = v₀² - 2 a x

When the speed stops it is zero

x₁ = v₀² / 2 a₁

Let's reduce speed

v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

x₁ = 80.667² / (2 11.2)

x₁ = 290.50 feet

The minimum braking distance

x₂ = 80.667² / (2 19.168)

x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

0 = v₀² - 2 a x

v₀ = √2 a x

We calculate the speed for the two accelerations

v₀₁ = √ (2 11.2 155)

v₀₁ = 58.92 feet / s

v₀₂ = √ (2 19.168 155)

v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago