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storchak [24]
3 years ago
12

A horizontal pull A pulls two wagons over a horizontal frictionless floor, the first wagon is 500N, the second is 2000 N. The te

nsion in the light horizontal rope connecting the wagons is.
help explain your answer,
Physics
1 answer:
MrMuchimi3 years ago
5 0
<span>The force will be zero if the wagons are moving at a constant speed (i.e. not accelerating), as there is no frictional force to overcome. If the wagons are accelerating, the force will be proportional to the acceleration, and 20% of the force applied by A.</span>
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To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
8 0
3 years ago
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
Which will a positively charged objects attract
Vitek1552 [10]
Me and you! enjoy your day girllllaa or boy
6 0
3 years ago
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What two measurements are needed to determine density?
g100num [7]
Density is defined as (mass) per unit (volume).  So in order to calculate
the density of a glob of some substance, you pretty much have to measure
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7 0
3 years ago
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The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a
Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

J=F.\Delta t

J=10^6\ N\times 20\ s

J=2\times 10^7\ N-s

(b) Impulse is also given by the change in momentum i.e.

J=\Delta p=p_f-p_i

J=p_f-p_i

p_f=J+p_i

p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s

p_f=20143000\ kg-m/s

(c) For new velocity,

v_f=\dfrac{p_f}{m}

v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}

v_f=1549.46\ m/s

Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
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