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2 years ago

In order for a living organism to substain its life it must be able to

1 answer:

8 0

Have the 6 characteristics of living things
1.movement/growth
2.reproduction
3.made of 1 or more cells
4.sensitivity
5.excretion
6.nutrition

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You are driving at 50 miles per hour. If you decrease the time it takes you to travel 1 mile by 8 seconds, what is your new spee

Dmitriy789 [7]

Answer:

The new speed is 56.25 miles/hour.

Explanation:

Since speed = distance/time;

time = distance/speed.

While driving at 50 miles/hour, time taken for one to complete 1 mile is (1/50) hour

(1/50) hour = (1/50) × 3600s = 72 seconds.

So, if this time to complete 1 mile (72 seconds) is reduced by 8 seconds,

New time to complete 1 mile will be = 72 - 8 = 64 seconds = (64/3600) hour = 0.0178 hour

New speed would be = (1 mile/64 seconds) = (1 mile/0.0178 hour) = 56.25 miles/hour.

Hope this Helps!!!

7 0

2 years ago

Which part of the eye holds eye color?

Anna [14]

Hi, it’s the iris. Hope this helped u

6 0

2 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

3 0

2 years ago

1200kg car speeds up from 2 m/s to 6 m/s. How many times greater is the car's kinetic

Bezzdna [24]

Answer:

9 times as great

Explanation:

KE = 1/2mv²

1/2(1200)(2)² = 2,400 J
1/2(1200)(6)² = 21,600 J

21,600 / 2,400 = 9

3 0

2 years ago

PLEASE ANSWER ASAP!! YOU WILL GET BRAINLIEST IF RIGHT!

Elena-2011 [213]

D just east of station b

6 0

2 years ago

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