Ask question

Ask question

All categories

2 years ago

9

A dog accelerates at 1.50 m/s2 to reach a velocity of 13.5 m/s while covering a distance of 49.3 m. What was his initial velocit

y?

1 answer:

vovangra [49]2 years ago

7 0

Let <em>v</em> be the dog's initial velocity. Then

(13.5 m/s)^2 - <em>v</em> ^2 = 2 (1.50 m/s^2) (49.3 m)

==> <em>v</em> ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)

==> <em>v</em> = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))

==> <em>v</em> ≈ 5.86 m/s

You might be interested in

Horatio drew a diagram to illustrate how magnetism is produced by electric currents

dimaraw [331]

<span>The magnetic field does not continually spread outward from the wire.</span>

4 0

3 years ago

Read 2 more answers

PLEASE SHOW STEPS!

SOVA2 [1]

Answer:

660 J/kg/°C

Explanation:

Heat lost by metal = heat gained by water

-m₁C₁ΔT₁ = m₂C₂ΔT₂

-(0.45 kg) C₁ (21°C − 80°C) = (0.70 kg) (4200 J/kg/°C) (21°C − 15°C)

C₁ = 660 J/kg/°C

8 0

3 years ago

Help please:Calculate the mechanical energy of a bird flying at a speed of 10 m / s at an altitude of 15 m. If its mass is 150 g

Sonja [21]

Mass of the bird(m) = 150 g = 0.15 kg

Speed (v) = 10 m/s

Kinetic Energy = $\frac{1}{2}m v^{2} = \frac{1}{2} 0.15 (10)^{2}$ = 7.5 J

Altitude (h) = 15 m

Gravitational Potential Energy = (0.15)(9.81)(15) = 22.0725 J

Mechanical Energy = Kinetic Energy + Potential Energy = 7.5 + 22.0725

= 29.5725 J

4 0

2 years ago

Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 4x + 4y + 4x2 + 4y2

maw [93]

Answer:

Q=185.84C

Explanation:

We have to take into account the integral

$Q=\int \rho dV$

In this case we have a superficial density in coordinate system.

Hence, we have for R: x2 + y2 ≤ 4

$Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx$

but, for symmetry:

$Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C$

HOPE THIS HELPS!!

8 0

2 years ago

A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa

Darya [45]

Answer:

x=0.53 $x10^{-3} m$

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1. $x10x^{-6} \frac{C}{m^{2}}$

ε=8.85 $x10^{-12} \frac{C^{2}}{n*m^{2}}$

$E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}$

Force of charge is

$F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N$

$F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s$

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

$x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2} \\x_{f}=0.53x^{-3}m$

5 0

2 years ago