A car accelerates from 6.0 m/s to 18 m/s at a rate of 3.0 m/s2. How far does it travel while accelerating

4. How does a molecule differ from an atom

tia_tia [17]

Answer:

Explanation:

In a molecule, atoms are bonded together by single, double, or triple bonds. An atom has a nucleus surrounded by electrons. ... So another difference between atoms and molecules is that when similar atoms combine together in varying numbers, molecules of different properties can be formed.

8 0

3 years ago

Read 2 more answers

On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m

nika2105 [10]

Answer:

The magnitude of the torque is 263.5 N.

Explanation:

Given that,

Applied force = 31 N

Distance from the axis = 8.5 m

She applies her force perpendicularly to a line drawn from the axis of rotation

So, The angle is 90°

We need to calculate the torque

Using formula of torque

$\tau=Fd\sin\theta$

Where, F = force

d = distance

Put the value into the formula

$\tau=31\times8.5\sin90$

$\tau= 263.5\ N$

Hence, The magnitude of the torque is 263.5 N.

4 0

2 years ago

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2. We want

lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

$\Delta x=v_ot+\dfrac{1}{2}at^2$

t is time taken by the ball to hit the ground

$v_o$ is initial speed of the ball

So, the correct option is (A).

8 0

3 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

3 years ago

A land rover drops from rest a vertical distance of 1000 m to the surface of planet in 15 seconds what is the magnitude of the a

umka21 [38]

Answer:

66.6 or it could be 66.66 one of those

Explanation:

6 0

3 years ago