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Savatey [412]
4 years ago
6

When driving at slower speeds you need to use what type of steering

Physics
1 answer:
OLga [1]4 years ago
8 0

Answer:

slower speeds = larger and faster steering wheel movements

faster speeds = small and slow steering wheel movements

Explanation:

When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.

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What acceleration occurs if the friction force in #4 is now 4N?
anastassius [24]

Answer:

you a b1tc h

Explanation:

you're not going anywhere in life. mcdonalds hiring $16 an hour where i live.

4 0
2 years ago
Read 2 more answers
Witch is the property of matter in witch a substance can transfer heat or electricity
saw5 [17]

Answer:

B. Conductivity.

Explanation:

Conductivity is the quantity of heat passing per second through a slab of unit cross-sectional area when the temperature gradient between the two faces is unity when put in heat.

4 0
4 years ago
An electromagnetic wave in a vacuum traveling in the +x direction generated by a variable source initially has a wavelength λ of
djverab [1.8K]

Answer:

B = E/c = 14.04T₁ = 11 pT

Explanation:

We know c = E/B where E = maximum electric field = 3.30 × 10⁻³ V/m, B = maximum magnetic field and c = speed of light

B = E/c also c = fλ = λ/T where λ = wavelength = 235 μm = 235 × 10⁻⁶ m and T = period

c = λ₁/T₁ = λ₂/T₂    T₂ = 2.8T₁ where λ₁,λ₂ are the initial and final wavelengths and T₁,T₂ are the initial and final periods.

T₁ =  λ₁/c = 235 × 10⁻⁶ m/3 × 10⁸ m/s = 7.833 × 10⁻¹³ s = 0.7833 ps

T₂ = 2.8T₁ = 2.8 × 7.833 × 10⁻¹³ s = 21.93 × 10⁻¹³ s = 2.193 ps

λ₁/T₁ = λ₂/2.8T₁

λ₂ = 2.8λ₁ = 2.8 × 235 μm = 658 μm

c = λ₂/T₂ = 2.8λ₁/2.8T₁ = λ₁/T₁ , since the speed of light c is constant.

B = E/c = E/λ₁/T₁ = ET₁/λ₁

B = ET₁/λ₁ = 3.30 × 10⁻³ V/m × T₁/235 × 10⁻⁶ m = 14.04T₁  Tesla

B = 14.04 × 7.833 × 10⁻¹³ s = 10.99 × 10⁻¹² T ≅ 11 pT

5 0
3 years ago
Will give correct answer brainliest
Grace [21]

Answer:

all will help you!!! !!!!!!!

7 0
3 years ago
An object moves from point A to B to C to D and finally to A
Nataly [62]

here's the solution,

The <em>radius</em> of the circle =<u> 3 km</u>

distance covered = <em>circumference</em> of the circle,

So, Circumference :

=》

2\pi r

=》

2 \times 3.14 \times 3

=》

18.84

(a). Distance covered by moving object is 18.84 km

(b). 0 km

now, Displacement of the object is 0 km, because displacement is the shortest distance from stating point to the destination, but the object returns back to the starting point, hence magnitude of displacement is 0.

5 0
3 years ago
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