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erastovalidia
1 day ago
12

When you’re driving on a level road and there’s ice on the pavement, you hardly notice that ice while you’re heading straight at

a constant speed. Why is it that you only notice how slippery the road is when you try to turn left or right, or to speed up or slow down?
Physics
1 answer:
Ostrovityanka [2.2K]1 day ago
3 0

Answer: Focusing solely on friction when accelerating or decelerating

Explanation:

This phenomenon arises because attempting to turn left or right involves changing the velocity of the vehicle, which in turn leads to acceleration and an ensuing alteration in force.

Alterations in force lead to a variation in friction, causing the road to feel slippery.

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Which trailer has more downward pressure where it attaches to the car?
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The trailer that is loaded the most. The total weight does not matter; it is about how the load is distributed. For example, our 12,000 lb snow cat trailer has weight distribution that results in less than 100 lbs of tongue weight. Heavy tongue weight can create issues, as it can shift the weight off the front wheels of the towing vehicle, causing instability.

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25 days ago
what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t
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Recall this formula for a device operating in a direct current circuit:
P = IV
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20 days ago
If the position of an object is zero at one instant, what is true about the velocity of that object?
ValentinkaMS [2425]
If the position of an object is zero at a particular moment, this does not provide any indication about its velocity. It might simply be moving through that point, and you observed it exactly when it was at zero.
6 0
29 days ago
3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta
ValentinkaMS [2425]

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

The performance of the heat pump can be calculated using the formula

β = TH / (TH - TC)

a.

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / (288.15 K - 253.15 K)

β = 8.23 K

b.

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / (288.15 K - 278.15 K)

β = 28.815 K

6 0
14 days ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Maru [2355]

Answer:

The final size is nearly the same as the initial size because the increase in size1.055\times 10^{- 7} is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, m_{p} = 1.67\times 10^{- 27} kg

Initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This proton energy is \simeq 250 GeV

Thus, the speed of the proton, v\simeq c

The time to cover 1 km = 1000 m of distance is calculated as:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

According to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
10 days ago
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