All categories

4 years ago

1) A train accelerates from 36 km/hr to 54 km/hr in 10 s. Find acceleration?

Physics

1 answer:

Crank4 years ago

3 0

Answer: Given:

Initial velocity= 36km/h=36x5/18=10m/s

Final velocity =54km/h=54x5/18=15m/s

Time =10sec

Acceleration = v-u/ t

=15-10/10=5/10=1/2=0.5 m/s2

Distance =s=?

From second equation of motion:

S=ut +1/2 at^2

=10*10+1/2*0.5*10*10

=100+25

=125m

So distance travelled 125m

Hope it helps you

You might be interested in

Which term is applied to an object through which light passes?

Elena L [17]

Answer:

Explanation:

transparent_objects that allows light to pass through and can you see through them

5 0

3 years ago

Read 2 more answers

What is the relation between centre of gravity and stability

maw [93]

Explanation:

tilting it will raise the height of its center of gravity.

8 0

3 years ago

Which choice can be correctly defined as “the sum of all charge in a closed system remains constant”?

Elena-2011 [213]

Answer:

Law conservation of charge (D)

Explanation:

8 0

3 years ago

Read 2 more answers

A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh

eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

$V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}$

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

$P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}$

$P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}$

$P^2 = \frac{P^2V^2}{c^2}+m^2V^2$

$P^2 = V^2 (\frac{P^2}{c^2}+m^2)$

$V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}$

$V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}$

$V = 2.81784*10^8m/s$

Therefore the height with respect the observer is

$l = l_0*\sqrt{1-\frac{V^2}{c^2}}$

$l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}$

$l = 0.56m$

Therefore the height which the observerd measure for her is 0.56m

8 0

3 years ago

if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use

julia-pushkina [17]

Answer:

$h_{moon} = 11.8\ m$

Explanation:

Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:

$W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} = 11.8\ m$

5 0

3 years ago