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3 years ago

The SI unit for acceleration is_____ a m/s. b m/s 2 c m/s2 d none of the above

Physics

1 answer:

zalisa [80]3 years ago

7 0

I dont know from option
Because SI Unit of acceleration is m/s^2

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A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Det

FromTheMoon [43]

Answer:

$a=2.5\ m/s^2$

Explanation:

Given that,

Initial speed of the car, u = 14 m/s

Finally, it comes to rest, v = 0

Time, t = 5.6 s

We need to find the average acceleration of the car during this time interval. We know that,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2$

So, the acceleration of the car is $2.5\ m/s^2$ in the opposite direction of motion.

3 0

3 years ago

The two rods are now firmly attached together as in Figure 2.

MakcuM [25]

Thermal equilibrium is attained and the both rods are now at the same temperature.

<h3>What is thermal equilibrium?</h3>

Two bodies are said to have attained thermal equilibrium when the two bodies at the same temperature. It should be known that when two rods are firmly attached to each other heat flows from one rod to another.

As such, after some time, thermal equilibrium is attained and the both rods are now at the same temperature.

Learn more about thermal equilibrium:brainly.com/question/2637015

4 0

2 years ago

If ram is 40kg and travels 200m in 30 sec find his power

WARRIOR [948]

Work done

N×200
mg200
40(10)(200)
400(200)
80000J

Power

Work done/Time
80000/30
8000/3
2668W

8 0

2 years ago

Read 2 more answers

A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu

Murljashka [212]

The work done on the ship is $1.5\cdot 10^7 J$

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we have:

$F=5.00\cdot 10^3 N$ (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

$\theta=0^{\circ}$ (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

$W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0

3 years ago

The relationship between a cathode and an anode involves

astra-53 [7]

It involves electrons.

The cathode is the electrode where electron deficient ions move to.

While the anode is electrode where electron excess ions move to.

So the relationship between Cathode and Anode involves electrons.

C.

5 0

3 years ago