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jeyben [28]
3 years ago
13

NEED Answers ASAP!!!!!

Physics
1 answer:
Maslowich3 years ago
6 0

The force on the truck is F=13650  N

<u>Explanation:</u>

Solving the problem

Given data,

M= 1500 kg

A= 9.1 m/s

We have the formula,

F= M×A

F= 1500 × 9.1

F=13650  N

The Force on the truck is F=13650  N

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In 1984 a team of german physicists synthesized a new element by smashing an iron-58 nucleus into a lead-208 target. if a neutro
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Here in all nuclear reactions we can say that mass conservation and charge conservation is always true

Here iron nuclei smashed into lead nuclei and then a new nuclei will form which will released along with a neutron

Now in this reaction mass and charge will remain conserved

mass number of iron + mass number of lead = mass number of new nuclei + mass number of neutron

58 + 208 = x + 1

x = 265

so the new nuclei formed will have mass number A = 265

now we will use charge conservation

Number of protons in iron + number of protons in lead = number of protons in new nuclei + number of protons in neutron

26 + 82 = z + 0

z = 108

so the new nuclei will form with atomic number z = 108 and mass number A = 265

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3 years ago
At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0 N/m is set into osci
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Answer:

Explanation:

Answer:

4.53 kg

Explanation:

spring constant, K = 16 N/m

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Maximum speed, v = 37.6 cm/s = 0.376 m/s

The formula for maximum speed is given by

v = ωA

where, ω is the angular frequency

0.376 = ω x 0.2

ω = 1.88 rad/s

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m = \frac{16}{1.88 ^{2}}

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Thus, the mass of banana bunch is 4.53 kg.

8 0
3 years ago
A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli
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Answer:

a.  v=3.11mls, 29.4^{0}

b.   K.E =-697.8J

Explanation:

To calculate the values in the  question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\

Momentum of player 2 p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\

Hence the total momentum p_{12}=p_{1}+p_{2}\\

Note, since the direction of movement before collision is due south and  due north respectively we have to represent the velocity using the rectangular coordinate

Hence  p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\

(95+90)v=475i+270j\\

v=2.57i+1.45j\\

solving for the resultant velocity, we have

v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls

To calculate the direction of movement, we have

\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\  \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\

And the final kinetic energy after collision is

K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\  K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J

The decrease in Kinetic energy is

K.E =K.E_{final}- K.E_{initial}=894.7-1592.5

K.E =-697.8J

The negative sign indicate a decrease in Kinetic energy

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