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nadya68
4 months 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? .
Physics
1 answer:
ValentinkaMS [3.4K]4 months 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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Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at 350°F, releasing a wonderful aroma
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The mass of the baked loaf will be lower than that of the dough.
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2 months ago
A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav
kicyunya [3294]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6 m/s^2

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 m/s^2.

   32 = 0 + 1.6 * t

    t = 20 seconds.

  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

6 0
3 months ago
Read 2 more answers
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
Softa [3030]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

\Phi =E\times S\times COS\Theta

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
3 months ago
A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution
Maru [3345]

Answer:

Explanation:

Amount of gold deposited = 0.5 g

Gold's molar mass = 197 g/mol

Time duration, t = 6 hours

= 6 × 3600

= 12600 s

Calculation of moles: mass/molar mass

= 0.5/197

= 0.00254 mole

Assuming

Au --> Au+ + e-

Faraday's constant = 9.65 x 10^4 C mol-1

Charge, Q = 96500 × 0.00254

= 244.924 C

Relation: Q = I × t

Thus, I = 244.924/12600

= 0.011 A

= 11.34 mA.

6 0
3 months ago
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