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ratelena
12 days ago
9

A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc

ity of 1 m/s north rela-tive to the ground. Which of the following is the best estimate of the velocity of the skateboard immediately after the dog has jumped?
(A) 1 m/s north(B) 1 m/s south(C) 3 m/s south (D) 7 m/s south(E) 17 m/s south
Physics
1 answer:
inna [3.1K]12 days ago
3 0

Answer:

17 m/s south

Explanation:

m_1 Mass of the dog = 10 kg

m_2 Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 Velocity of the dog = 1 m/s

u_2 Velocity of skateboard

In this scenario, linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

The speed of the skateboard post-dog jump will be 17 m/s south, considering north as the positive direction

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Softa [3030]

Answer:

3.4 x 10⁴ m/s

Explanation:

Analyze the circular path of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of an electron = 9.1 x 10⁻³¹ kg

v  = speed in the radial direction

r = radius of the circular trajectory = 2 mm = 0.002 m

q = charge of an electron = 1.6 x 10⁻¹⁹ C

For the electron’s circular movement

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Now, consider the electron's movement in a straight line:

v' = speed in linear motion

x = distance traveled horizontally = 9 mm = 0.009 m

t = duration = \frac{2\pi m}{qB} = \frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})} = 4.5 x 10⁻⁷ sec

Using the formula

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

The resultant speed is given by

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s

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An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and
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The electric flux through the cylindrical surface surrounding the infinite charged wire is given by the formula ∅E = E x 2πrl. To analyze this, we consider an infinitely long straight wire with a uniform linear charge density of λ Cm⁻¹. The electric field at a distance r from this charge can be evaluated using a cylindrical Gaussian surface of radius r and length l, oriented along the wire. Only the curved surface of the cylinder contributes to the total flux since the other surfaces are perpendicular to E.
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8 days ago
An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
Ostrovityanka [3204]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time t_{sound}. Thus, we can derive:

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

t_{arrow} + t _{sound} = 1 s.

Using this relationship in the distance formula for sound allows us to write:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Substituting the value of d from the first equation yields:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, after some calculations, we can proceed further:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Finally, the value is inserted into the initial equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

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26 days ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Sav [3153]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

The question indicates that

    The wire’s diameter is  d = 0.205cm = 0.00205 \ m

     The radius of the wire is  r = \frac{0.00205}{2} = 0.001025 \ m

     Aluminum's resistivity is 2.75*10^{-8} \ ohm-meters.

       The electric field variation is described as

         E (t) = 0.0004t^2 - 0.0001 +0.0004

     

The charge is effectively given by the equation

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area expressed as

       A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 Thus,

       \frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

By substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.

The question states that t =  5 seconds

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
1 month ago
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