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12 days ago

9

If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti

ve period? .

1 answer:

ValentinkaMS [1.1K]12 days ago

8 0

Speed is defined as distance over time. Hence, to determine the distance, we use d = V * t. Plugging in the values yields d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km. Thus, during this distracted period, a distance of 0.08Km was covered.

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Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

Ostrovityanka [942]

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14 days ago

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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

inna [987]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ $\frac{du}{dy}$

therefore

= µ $\frac{v}{h}$ ............1

substituting the values

= 1.75 × $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

the area between the air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

now applying Newton's second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

solving for t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

the time needed after impact for the puck is 2.18 seconds

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6 days ago

On its own, a certain tow-truck has a maximum acceleration of 3.0 m/s2. what would be the maximum acceleration when this truck w

inna [987]

Let us denote $a_1=3 m/s^2$ as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is
$F= ma_1$
where m is the mass of the truck alone.

Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is
$F=3m a_2$
where a2 represents the new acceleration.
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:
$m a_1 = 3 m a_2$
and solving for a2 gives us:
$a_2 = \frac{m a_1}{3 m}= \frac{a_1}{3}= \frac{3 m/s^2}{3}=1 m/s^2$

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12 days ago

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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

Softa [913]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Charge (Q₁) = 572.8 $* 10^{-9}$ C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² = 0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

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4 days ago

A car rolls down a ramp in a parking garage. The horizontal position of the car in meters over time is shown below. Graph of ver

Ostrovityanka [942]

Answer:

d_total = 12 m

Explanation:

In this kinematics scenario illustrated in the graph provided, we determine the distance traveled over a 24-second duration.

The comprehensive distance can be calculated as follows:

d_total = d₁ + d₂ + d₃

Given that d₂ on the graph is level (v=0), its distance equates to zero, hence d₂ = 0.

The distance for d₁ is calculated as:

d₁ = 12 - 6 = 6 m

For distance d₃:

d₃ = 6 - 0 = 6 m

Thus, the overall distance covered is:

d_total = 6 + 0 + 6

d_total = 12 m

3 0

2 days ago