Ask question

Ask question

All categories

VashaNatasha [74]

3 years ago

15

a bus starts from rest and attains 40 kilometres per hour velocity after 10 seconds calculate the acceleration of the bus in km

-2 and ms -2

1 answer:

igomit [66]3 years ago

8 0

Answer:

Acceleration = Final velocity-initial velocity/time

In m/s^2

a = 40000-0 /10

= 4000m/s

in km/h

a = 40/10÷360

= 1440km/h

You might be interested in

Explain the difference between what Isaac Newton and Louis de Broglie would have to say about the momentum of a particle that is

pishuonlain [190]

Answer:

Explained

Explanation:

Newton would resort to the classical mechanics and say that the momentum of the particle that is moving with a constant velocity will be given by: momentum = mass x velocity

this approach will highlight the particle nature and will not be relativistic.

De-Broglie will say that the momentum of the particle is related to its associated matter wave and the relation between them is given by:

$p = \frac{h}{\lambda}$

where \lambda = wavelength of the matter wave associated to the particle, h = planck's constant

and $p = \gamma\times mv$

thus, this highlights the wave nature of the particle and is also relativistic.

6 0

3 years ago

A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

8 0

3 years ago

Can someone help asap with this

kondaur [170]

Answer:

B

Explanation:

i had a test on this and got it correct

7 0

3 years ago

A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction

Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed = 25.5m/s

Acceleration = 1.94m/s²

Time = 2.3s

Unknown:

Final speed of the car = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

v = u + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

Now insert the parameters and solve;

v = 25.5 + (1.94 x 2.3) = 29.96m/s

3 0

3 years ago

A toaster draws 8 A of current with a voltage of 120 V. Which is the power used by the toaster?

dexar [7]

$U=120 \text{ V}\\
I=8\text{ A}\\
P=U\cdot I\\\\
P=120\cdot8=960 \text{ W}
$

7 0

3 years ago