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3 years ago

How is a controlled variable different from a responding variable?

Physics

1 answer:

Arisa [49]3 years ago

5 0

Answer:

The answer is a controlled variable stays the same throughout an experiment, but a responding variable changes

Explanation:

just did it on apex

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A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted

Tpy6a [65]

Answer:

The normal force the seat exerted on the driver is 125 N.

Explanation:

Given;

mass of the car, m = 2000 kg

speed of the car, u = 100 km/h = 27.78 m/s

radius of curvature of the hill, r = 100 m

mass of the driver, = 60 kg

The centripetal force of the driver at top of the hill is given as;

$F_c = F_g - F_N$

where;

Fc is the centripetal force

$F_g$ is downward force due to weight of the driver

$F_N$ is upward or normal force on the drive

$F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N$

Therefore, the normal force the seat exerted on the driver is 125 N.

6 0

2 years ago

Explain using physics vocabulary WHY the volume increases as the temperature increases.

Sveta_85 [38]

Answer:

P V = n R T ideal gas equation

V = k T where k = a constant and equals k = n R / P

V is proportional to T when other factors are constant

4 0

2 years ago

What is the name for a wave that requires energy to produce it and a medium to travel through?

Wittaler [7]

....mechanical wave...

8 0

3 years ago

The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0

2 years ago

A tennis player strikes a tennis ball from underneath with her racket. The ball is sent straight up with an initial velocity of

Stels [109]

So the acceleration of gravity is 9.8 m/s so that’s how quickly it will accelerate downwards. You can use a kinematic equation to determine your answer. We know that initial velocity was 19 m/s, final velocity must be 0 m/s because it’s at the very top, and the acceleration is -9.8 m/s. You can then use this equation:

Vf^2=Vo^2+2ax

Plugging in values:

361=19.6x

X=18 m

6 0

2 years ago