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Rus_ich [418]
3 years ago
15

You are riding your bike at 9 km/h twn minutes later your speed is 6 km/h

Physics
2 answers:
spayn [35]3 years ago
7 0

Your average acceleration is

Negative 1.5 km/hr per minute .

gtnhenbr [62]3 years ago
5 0

De-celleration because speed is lower as time goes on


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What is the Ozone layer?
yaroslaw [1]
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4 0
3 years ago
Read 2 more answers
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu
tensa zangetsu [6.8K]

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

\Delta V = E d

where

\Delta V is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

E=5.00\cdot 10^6 V/m

d = 2.50 mm = 2.50\cdot 10^{-3} m

Substituting, we find

\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V

3 0
3 years ago
A geologist is trying to determine what time period a layer of rock was formed.
vodka [1.7K]

Answer:

B

Explanation:

You always want to test as many samples as possible

8 0
3 years ago
Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the tex
kompoz [17]

Answer:

L = 0.0319 H

Explanation:

Given that,

Number of loops in the solenoid, N = 900

Radius of the wire, r = 3 cm = 0.03 m

Length of the rod, l = 9 cm = 0.09 m

To find,

Self inductance in the solenoid

Solution,

The expression for the self inductance of the solenoid is given by :

L=\dfrac{\mu_o N^2 A}{l}

L=\dfrac{4\pi\times10^{-7}\times(900)^{2}\times\pi(0.03)^{2}}{0.09}

L = 0.0319 H

So, the self inductance of the solenoid is 0.0319 henries.

4 0
3 years ago
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