All categories

3 years ago

A car speeds up from rest to +16 m/ s in 4s. calculate the acceleration

2 answers:

6 0

The magnitude of acceleration is (change in speed) / (time for the change).

Change in speed = (speed at the end) - (speed at the beginning) =

(16 m/s) - (0) = 16 m/s .

Time for the change = 4 s .

Magnitude of acceleration = (16 m/s) / (4 s) = 4 m/s per sec = 4 m/s² .

attashe74 [19]3 years ago

5 0

16 m/s - 0 m/s divided by 4s
= 16 m/s divided by 4s
= 4 m/s
The acceleration is 4 m/s squared

You might be interested in

Please help me with my quiz

Oduvanchick [21]

Answer:

matter: A

data: E

variable: C

Controlled: D

Physical: B

Explanation:

8 0

2 years ago

HELP I NEED THE ANSWER NOW

Setler79 [48]

Answer:

Speed of light

Explanation:

The famous Einstein's equation is:

$E=mc^2$

where

E is the energy

m is the mass

$c=3\cdot 10^8 m/s$ is the speed of light

In this equation, Einstein summarized the following fact: mass can be converted into energy, and the amount of energy released in such a process is given by the equation.

An example of application of this equation is the nuclear fusion process. In a nuclear fusion, two lighter nuclei combine into a heavier nucleus. However, the mass of the heavier nucleus is slightly less than the sum of the masses of the two original nuclei: some of the mass of the original nuclei has been converted into energy, accorging to the previous equation.

4 0

3 years ago

Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

2 years ago

The diagram shows a ballistic pendulum. A 200 g bullet is fired into the suspended 4 kg block of wood and remains embedded insid

adoni [48]

Question

What was the initial momentum of the bullet before collision?

Answer:

10 Kg.m/s

Explanation:

Momentum is a product of velocity of an object in m/s and its mass in kgs hence numerically expressed as p=mv where p is momentum, v is velocity and m is mass. Substituting m for 0.2 kg and v for 50 m/s then p=0.2*50=10 kg.m/s

5 0

3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago