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Afina-wow [57]
3 years ago
7

On a straight road with the +x axis chosen to point in the direction of motion, you drive for 5 hours at a constant 20 miles per

hour, then in a few seconds, you speed up to 60 miles per hour and drive at this speed for 1 hour.
What was the x component of average velocity for the 6-hour period, using the fundamental definition of average velocity, which is the displacement divided by the time interval?
Physics
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

v = 26.7 mph

Explanation:

During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:

Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi

Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:

Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi

So, the total displacement during all the trip wil be:

Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi

So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:

v = Δx / Δt = 160 mi / 6 h = 26.7 mph

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100mL of 4°C water is heated to 37 °C . Assume the density of the water is 1g/mL. The specific heat of water is 4.18 J/g(°C). Wh
swat32

Answer:

13807.2  J/g°C

Explanation:

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6 0
3 years ago
On a very muddy football field, a 110kg linebacker tackles an 85kg halfback. Immediately before the collision, the linebacker is
Alchen [17]
<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback

So for conservation of momentum,
rho = mv

M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf

For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2

Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
6 0
3 years ago
Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4
arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

  • initial velocity of the car, u = 76 m/s
  • acceleration of the car, a = - 9 m/s²
  • time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

  • height above the ground, h = 500 m
  • velocity of spaceship, u = 150 m/s
  • time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) -  (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m

Learn more here: brainly.com/question/24527971

5 0
2 years ago
Diamond shine brightly but the pieces of glass don't​
icang [17]

Answer:

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Explanation:

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5 0
2 years ago
An electron is projected with an initial speed v0 = 1.10 x 10⁶ m/s into the uniform field between the parallel plates. The dista
ArbitrLikvidat [17]

Answer:

a) E=364N/C

b) No

Explanation:

A) Because the electron is affected by an acceleration force in this case by the electric field, we can use the formulas of 2-dimension movement.

We will assume the electron missed the upper plate, so we need to calculate the time to travel all the way through the plate, that is:

x=v_x*t

where:\\x=distance\\v=speed\\t=time

so:

t=\frac{x}{v_x}=\frac{0.02m}{1.6\cdot 10^6m/s}\\\\t=1.25\cdot10^{-8}

the electron experiences an accelerated motion in the vertical direction, so we can obtain the acceleration of the electron:

y=\frac{1}{2}.a.t^2\\\\where:\\y=vertical\_distance\\a=acceleration\\t=time

so:

a=\frac{2.y}{t^2}\\\\a=\frac{2*(\frac{0.01}{2}m)}{(1.25\cdot10^{-8}s)^2}\\\\a=6.4\cdot10^{13} m/s^2

now we can use the relation:

F=m.a=E.q\\so\\E=\frac{m.a}{q}

where:\\\\E=electric\_field\\m=electron\_mass=9.1\cdot10^{-31}kg\\q=Charge=1.6\cdot10^{-19}\\a=acceleration

Now we can calculate the electric field:

E=\frac{9.1\cdot10^{-31}kg\cdot6.4\cdot10^{13}m/s^2}{1.6\cdot10^{-19}C}\\\\E=364N/C

B) Because the proton has the same charge but positive it will go in the opposite direction, so because we assume the electron didn't touch the plate, the proton won't.

5 0
3 years ago
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