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2 years ago

12

4: Question 2 - 10.0 pts possible

1 answer:

Orlov [11]2 years ago

8 0

Answer:

5. All of the statements are true; non is false

Explanation:

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An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travel

Licemer1 [7]

Answer:

$v=4.44\frac{m}{s}$

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero) and accelerates at a constant rate, position can be described like this: $x=v_{0}t +\frac{1}{2} at^{2}$ where x is the position, t is the time a is the acceleration and $v_{0}$ is initial velocity. In this way acceleration can be found. $a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }$ .

Now we are able to found velocity at any time with the formula: $v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}$

3 0

3 years ago

Which is not a device for reproducing sound?

Alona [7]

Phonograph. Hope that helps

7 0

2 years ago

What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between the

AlladinOne [14]

Answer:

2.87m

Explanation:

Using the law of gravitation to solve this question

F = GMm/r²

G is the gravitational constant

M and m are the masses

r is the distance between the masses

Substitute the given values

G = 6.67×10^-11 m³/kgs²

M =8.8 x 10^6 kg

m = 5.6 x 10^5 kg

F =440N

400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²

400r² = 328.698×10

400r² = 3286.98

r² = 3286.98/400

r² = 8.21745

r = √8.21745

r = 2.87m

Hence the distance of separation is 2.87m

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2 years ago

Fill in the blank for four and five answer. Question for number 6.

ehidna [41]

Increase .... decrease .... presumably it's the "best shape" for a body which has been formed by the gravitational force

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2 years ago

At a race car driving event, a staff member notices that the skid marks left by the race car are

krok68 [10]

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

$v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s$

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

5 0

2 years ago