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denis-greek [22]
2 years ago
9

Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain

how the following will be different for car 2: a) braking distance, b) thinking distance and c) stopping distance.
Physics
1 answer:
Zepler [3.9K]2 years ago
8 0

Answer: car 1 is going how fast

Explanation: no need to answer without speed I won't know distance.

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Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?
cricket20 [7]

Answer;

B. Fluorescent lamps operate at a higher temperature than incandescent lamps.

Explanation;

-A fluorescent lamp, is a type of electric light (lamp) that uses ultraviolet emitted by mercury vapor to excite a phosphor, which emits visible light.

-A fluorescent lamp produces less heat, thus, it is much more efficient. A fluorescent bulb can produce between 50 and 100 lumens per watt. This makes fluorescent bulbs four to six times more efficient than incandescent bulbs.

-Fluorescent lamps operate best around room temperature. At much lower or higher temperatures, efficacy decreases.

4 0
2 years ago
Read 2 more answers
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
Part A
7nadin3 [17]

Answer:

2.5 m/s²

Explanation:

Using the formula, v = u + at ( v = Final velocity; u = Initial velocity; t = Time; a = Acceleration)

25 = 0 + 10a

a = 25/10 = 2.5 m/s²

8 0
3 years ago
A spring with spring-constant of 100 n/m is compressed 0.10 m. what is its maximum stored elastic potential energy
Ganezh [65]

Elastic potential energy stored in a spring is

(1/2) · (spring constant) · (stretch or compress)² .

PE = (1/2) · (100 N/m) · (0.1 m)²

PE = (50 N/m) · (0.01 m²)

PE = (50 · 0.01) (N · m / m²)

PE = 0.5 N · m

PE = 0.5 Joule

5 0
3 years ago
In a series circuit, when batteries are added to the circuit the voltage will _____ and the current will _____.
Eva8 [605]

Answer: voltage, explode is the answers

Explanation:

Bc of the device for each  server

5 0
3 years ago
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