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4 years ago

What are the three reasons cells divide?

1 answer:

7 0

1 growth. Go from one cell/( zygote to a trillion)
2 replace. Repair\ 50 million cells die second.
3 reproduction. ( make cells for reproduction make specialized sex cells)

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There is no question my bad

Llana [10]

Answer:

It's cool dude cause more points for me

8 0

3 years ago

If 4.5×105kg of emergency cooling water at 10 ∘C are dumped into a malfunctioning nuclear reactor whose core is producing energy

Stella [2.4K]

Answer:

$\Delta t= 2962.395\,s\,(49.373\,min)$

Explanation:

Let assume that cooling water works under a pressure of 1 atmosphere. The time required to boil half of the water is determined by the First Law of Thermodynamics:

$\dot Q \cdot \Delta t = m \cdot [c_{p,w}\cdot (T_{2}-T_{1})+h_{fg}]$

$\Delta t = \frac{m\cdot [c_{p,w}\cdot (T_{2}-T_{1})+h_{fg}]}{\dot Q}$

$\Delta t = \frac{(2.25\times 10^{5}\,kg)\left[\left(4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (100\,^{\textdegree}C - 10\,^{\textdegree}C)+2256.5\,\frac{kJ}{kg} \right]}{200000\,kW}$

$\Delta t= 2962.395\,s\,(49.373\,min)$

8 0

4 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

Bingel [31]

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

distance travel in the reaction time

d₁ = v x t

d₁ = 20 x 0.5 = 10 m

distance travel after you apply brake

using equation of motion

v² = u² + 2 a s

0 = 20² - 2 x 10 x s

s = 20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

distance between car and the deer = 38 m - 30 m

= 8 m

b) now, maximum speed car.

distance travel in reaction time

d₁ = s x t

d₁ = 0.5 V

distance left between them

d₂ = 38 - d₁

d₂ = 38 - 0.5 V

distance travel after you apply brake

using equation of motion

v² = u² + 2 a d₂

0 = (V)² - 2 x 10 x (38 - 0.5 V)

V² + 10 V - 760 = 0

now, solving the quadratic equation

$x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}$

V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

7 0

3 years ago

A ball with a mass of 1 kg is moving in a straight line at the same speed as a ball with a mass of 10 kg. Both balls are brought

Alenkasestr [34]

Answer:

A it takes less force to stop the 1 kg ball because it has less inerita

Explanation:

7 0

3 years ago

Read 2 more answers

8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

nasty-shy [4]

Answer: The ball is travelling with a speed of 5.5 m/s after hitting the bottle.

Explanation:

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

$(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s$

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0

3 years ago