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2 years ago

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A person drives an automobile with a mass of 450 kilograms at a velocity of 26 meters per second. The driver accelerates to a ve

locity of 30 meters per second. The difference in the automobile’s kinetic energy between the two velocities is joules. You may use the calculator.

1 answer:

gregori [183]2 years ago

7 0

Ec₁=mv₁²/2=450*26²/2=225*676=152100 J

Ec₂=mv₂²/2=450*30²/2=225*900=202500 J

ΔEc=Ec₂-Ec₁=202500-152100=50.4 kJ

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Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

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2 years ago

A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

4 0

2 years ago

The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time

Natali5045456 [20]

<span>x=((12.3/100)m)cos[(1.26s^−1)t]
v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
v=<span> <span>-0.13261622 m/s
</span></span>the object moving at 0.13 m/s <span>at time t=0.815 s</span>

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2 years ago

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

Mandarinka [93]

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 = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

2 years ago

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.91 J1.91 J of work to set the ball s

Black_prince [1.1K]

Answer:

ω = 4.07 rad/s

Explanation:

By conservation of the energy:

W = ΔK

$1.91J = I/2*\omega^2$

where $I = 2/3*m*R^2=0.23kg.m^2$

Solving for ω:

$\omega = \sqrt{W*2/I} =4.07rad/s$

7 0

2 years ago