The sentence Walking the wire, my dog waited for the squirrel to fall is an example of _____.split infinitive
parallel structure
dangling participle
none of these

Answers

Answer 1

Answer: The answer is dangling participle. The dog wasn't walking the wire, the squirrel was. It is modifying the wrong noun.

Answer 2

Answer:

Answer:

dangling participle

Explanation:

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A person 4ft tall casts a shadow 10ft long. at the same time, a nearby tree casts a shadow 23 ft long. find the height of the tree

Use your understanding of d-t graphs to describe what is happening in the graph.

Answers

The speed changes after the point. C’mon Dude this is so easy

Support that P(e)=0.76,P(F)=0.45,and P(E and F)=0.32. What is P (E or F)?

Answers

The probability of (E and F) = P(e) times P(f) = 0.76 x 0.45 = 0.342 .

The probability of (e OR f) =

   1 - P(not e AND not f) =

   1 - P(not e) x P(not f) =

   1 - (1 - 0.76) x (1 - 0.32) =

   1 - ( 0.24 )   x   ( 0.68 ) = 1 - 0.1632 = 0.8368.

A car accelerates from 25 km/hr in 30 seconds. what Is the acceleration?

Answers

25 km/hr = 6,9 m/s
acceleration = 6,9 / 30 = 0,23

acceleration = 0,23 m / s²

acceleration is units of distance per time squared, take speed divided by time:

(25 km/hr)/(30 sec) = 0.833 km/hr/sec

these units aren't very helpful, so convert to units more likely to represent what your measuring. probably meters per seconds squared:

(0.833 km/hr/sec)*(1 hr/ 3600sec)*(1000m/km) = 0.23 m/s^2

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h, determine its minimum stopping distance (in ft), assuming the same rate of acceleration.

Answers

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

$v^2 = u^2 + 2as\n\nv = 0 mi/h\n\nu = 44 mi/h\n\ns = 0.00833 mi\n\n=> 0^2 = 44^2 + 2 * a * 0.00833\n\n=> 1936 = -0.01666a\n\na = -116206.48 mi/h^2 or -14.43 m/s^2$

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

$v^2 = u^2 + 2as\n\nv = 0 mi/h\n\nu = 79 mi/h\n\na = -116206.48 mi/h\n\n=> 0^2 = 79^2 + (2 * -116206.48 * s)\n\n6241 = 232412.96s\n\ns = (6241)/(232412.96) \n\ns = 0.0268531 mi = 141.78 ft$

The minimum stopping distance is 141.78 ft.

Gravity is a force of attraction between objects that is due to their weight and height between one another. a. True
b. False

Answers

The statement above is false. The statement should be: "Gravity is a force of attraction between objects that is due to their masses." Mass and weight are often interchanged. Gravitational force depends on the masses of objects, but the size of this force is what weight is. So the relationship is:

Weight = mass x gravitational constant = mass x 9.81 m/s^2

Mix 4 L of 30°C water is 6 L of 40°C water and you’ll have water at what temperature?