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Zolol [24]
4 years ago
7

A car is travelling north at a constant speed of 20 m/s. A force is applied directly east on the car. The motion of the car will

be change in which of these ways? A) The speed of the car will increase. B) The speed of the car will decrease. C) The speed will increase, and the car will change the direction it is moving. D) The speed will remain the same, but the car will change the direction it is moving.
Physics
1 answer:
Darina [25.2K]4 years ago
3 0

Answer:

C) The speed will increase, and the car will change the direction it is moving.

Explanation:

The car is travelling north at constant speed.  The force applied to the car is in the east direction, so it won't affect the speed in the north direction, but it will increase the speed in the east direction.  Therefore, the net speed increases, and the car changes direction.

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The emt must assume that any unwitnessed water-related incident is accompanied by:________
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3 0
2 years ago
Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe
NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B )  Current required through this wire

= 12 / .084 A

= 142.85 A

C )

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resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

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6 0
3 years ago
Read 2 more answers
A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.4 106 m/s.(a) Find the
nata0808 [166]

Answer:

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

c)  1.70×10⁻¹¹m

d) 1.661×10⁻²⁷KJ

Explanation:

A proton in the field experience a downward force of magnitude,

F = eE. The force of gravity on the proton will be negligible compared to the electric force

F = eE

a= eE/m

= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷

= 5.851× 10¹⁰m/s²

b)

V = u + at

u= 0

v= 1.4106m/s

v= (0)t + at

t= v/a

= 1.4106m/s/5.851 ×10¹⁰

= 2.411×10⁻¹¹s

c)

S = ut + at²

= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²

= 1.70×10⁻¹¹m

d)

Ke = 1/2mv²

    = (1.67×10⁻²⁷×)(1.4106)²/2

 =  1.661×10⁻²⁷KJ

5 0
3 years ago
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