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Force and Motions: Pushes and pulls of all kinds are________.

Mrrafil [7]

Pushes and pulls are all kinds of force.

5 0

3 years ago

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Orion *

nata0808 [166]

The answer is a) the hunter

8 0

3 years ago

What is the average speed of an object that travels 50 meters in 2 seconds? Select one: a. 100 m/s b. 25 m/s c. 50 m/s d. 2 m/s

AURORKA [14]

Answer:

Average speed, v = 25 m/s

Explanation:

It is given that,

Distance travelled by the object, d = 50 meters

Time taken, t = 2 seconds

Average speed is defined as the total distance divided by total time taken i.e.

$v=\dfrac{d}{t}$

$v=\dfrac{50\ m}{2\ s}$

v = 25 m/s

So, the sped of the object is 25 m/s. Hence, this is the required solution.

8 0

4 years ago

How do oxygen and beryllium atoms transform into oxygen ion O2- and Be2 beryllium ion Be2? tysm

Hoochie [10]

On the whole, the metals burn in oxygen to form a simple metal oxide. Beryllium is reluctant to burn unless it is in the form of dust or powder. Beryllium has a very strong (but very thin) layer of beryllium oxide on its surface, and this prevents any new oxygen getting at the underlying beryllium to react with it.

7 0

3 years ago

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Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia

Sedaia [141]

Answer:

$a_2\ =\ -33.65\ m/s^2$

Explanation:

Given,

For the first rocket,

Initial velocity of the first rocket A = $u_1\ =\ 4600\ m/s.$

Acceleration of the first rocket = $a_1\ =\ -18\ m/s^2$

For the second rocket,

Initial velocity of the second rocket B = $u_2\ =\ 8200 m/s.$
Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

$\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec$

Let $a_2$ be the acceleration of the second rocket B for the same time interval

from the kinematics,

$\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\$

$\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2$

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

6 0

3 years ago