The amount of electrons that an atom loses, shares or gains is the _______________

a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of

Papessa [141]

The initial speed of the bolt is not 58.86 m/s.

Let a be the acceleration of the rocket.

During the 4 sec lift off, the rocket has reached a height of

h = (1/2)*a*t^2

with t=4,

h = (1/2)*a^16

h = 8*a

Its velocity at 4 sec is

v = t*a

v = 4*a

The initial velocity of the bolt is thus 4*a.

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,

h = (1/2)*g*t^2 + V0*t

Substituting h0=8*a, t=6 and V0=-4*a into it,

8*a = (1/2)*g*36 - 4*a*6

Solving for a

a = 5.52 m/s^2

6 0

2 years ago

Cassie lives two miles away from school. Which mode of transport could benefit her health?

lord [1]

B.) Riding a bicycle to school.

4 0

3 years ago

Read 2 more answers

On a cold winter day, the flow of heat is from the outside in. A.True B.False

fredd [130]

Your answer s obviously false

8 0

3 years ago

Read 2 more answers

1. Black Panther ran 567.5 ks to the east in 2.3 hours then tumed around and ran 2218 km to

mario62 [17]

Answer:Average Velocity of the trip=98.77km/h

Explanation:

Given that:

Black Panther started his journey with

Distance ran towards east = 567.5km

Time ran towards east = 2.3 hours

and then turned back,

Distance ran towards west, = 2218 km

Time taken towards west, = 1.2 hours

Let us treat the westward movement as negative movement of the east

So total displacement =567.5 km east + (-2218km east )= 1,650.5km

But it seems the value of the distance ran west is wrong , because i do not think he can run 2218km under 1.2hours considering he ran 567.5m due east at 2.3hours .

So Let we use 221.8 km as distance ran due west.

So that Total Displacement becomes = 567.5 km east + (-221.8km east )= 345.7km

Total time = 2.3 + 1.2 = 3.5hrs.

Average velocity = Total displacement / total time

.345.7km / 3.5hr = 98.77km/h

7 0

3 years ago

What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

$\rho_w =1 g/mL \rightarrow \text{Water Density}$

$\rho_o = 0.82g/mL \rightarrow \text{Object density}$

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

$\sum F= 0$