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3 years ago

PLZZZ HELPP ASAP!!!!

Physics

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

1.68N

Explanation:

W=F*d

KE=1/2*M*V^2

M=32kg

V= 5.1m/s

D=50m

0.5(32)(5.1^2)= 416 J

Initial energy=500 J

final energy= 416 J

lost energy= 500-416

heat lost to friction=84 J

84=F*50

84/50=F*50/50

F=1.68N

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

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Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

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3 years ago

Coach Hogue is riding his motorcycle in a circle on wet pavement. Suddenly the bike slides out from under him. What failed to pr

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Answer:

The correct option is;

Force of Friction

Explanation:

As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path

The motion equation is

v = ωr and we have the centripetal acceleration given by

α = ω²r and therefore centripetal force is then

m×α = m × ω²r = m × v²/r

The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.

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3 years ago

A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m

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Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

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ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

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y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

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y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

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