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Iteru [2.4K]
3 years ago
14

A moving cyclist sees traffic up ahead and begins slowing down with constant acceleration. After slowing down for 0.5\,\text{s}0

.5s0, point, 5, start text, s, end text, the cyclist has travelled forward 2\,\text m2m2, start text, m, end text and has a speed of 10\,\dfrac{\text{km}}{\text{h}}10 h km ​ 10, start fraction, start text, k, m, end text, divided by, start text, h, end text, end fraction. We want to find the acceleration of the cyclist while slowing down to 10\,\dfrac{\text{km}}{\text{h}}10 h km ​ 10, start fraction, start text, k, m, end text, divided by, start text, h, end text, end fraction.
Physics
1 answer:
Helen [10]3 years ago
8 0

Answer:

Δx=vt−21​at2

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the acceleration as the cyclist slows down.

Assuming the initial direction of travel is the positive direction, our known variables are

t=0.5s

v=10km/h

Δx=2m

Since we don't know the initial velocity (Vo), and we are not asked to find (Vo)  we could use the kinematic formula that is missing (Vo) to solve for the target unknown, A

Δx=vt−21​at

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Where did the missouri compromise imaginary line run
lara31 [8.8K]

In the Missouri Compromise, the slavery line for future US states ran along the southern border of Missouri at 36 degrees north 30 minutes

7 0
3 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
Which electromagnetic waves have the shortest wavelength and the highest frequency?<br><br> waves
ANTONII [103]

Answer:

The correct answer is Gamma Rays.

Explanation:

5 0
3 years ago
Read 2 more answers
A liquid of density 1288 kg/m3 flows with speed 2.88m/s into a pipe of diameter 0.24 m. The diameter of the pipe decreases to 0.
Tju [1.3M]

Answer:

66.35m/s

Explanation:

Para resolver el ejercicio es necesario la aplicación de las ecuaciones de continuidad, que expresan que

A_1V_1 =A_2 V_2

From our given data we can lower than:

R_i = \frac{0.24}{2} = 0.12m

R_f = \frac{0.05}{2} = 0.025m

So using the continuity equation we have

A_1V_1 =A_2 V_2

V_2 = \frac{A_1V_1}{A_2}

V_2 = \frac{(\pi(0.12^2))(2.88)}{(\pi (0.25)^2)}

V_2 = 66.35m/s

Therefore the velocity at the exit end is  66.35m/s

8 0
3 years ago
6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final
Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy =  ¹/₂m₁u₁² + ¹/₂m₂u₂²  = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

                                                               = ¹/₂ x (9.43)²(25+6.8)

                                                               = 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

                               

4 0
3 years ago
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