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3 years ago

If a body is thrown up with an initial velocity 'u' and covers a maximum height of 'h' , then 'h' is equal to u²/2g

Physics

2 answers:

umka21 [38]3 years ago

5 0

GIVEN:

s = h , v = 0m/s and a = - g

ANSWER:

We know the equation, v² - u² = 2as.

Hence, - u² = - 2gh

so, maximum height attained is

h = u²/2g.

baherus [9]3 years ago

4 0

Answer:

1/2 m u^2 initial Kinetic Energy of body

m g h Potential Energy body attains as it rises to height

1/2 m u^2 = m g h conversion of KE to PE

h = u^2 / ( 2 g)

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2. Which of the following is an example of work being done on an object? A prism scatters ultraviolet light into visible light.

Liula [17]

A man pushes a couch across the room is the answer!

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3 years ago

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A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t

Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

$a=\dfrac{2\times 12\ m}{(1.2\ s)^2}$

a = 16.67 m/s²

Now put the value of a in equation (1) as :

$q=\dfrac{ma}{E}$

$q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}$

q = 0.0000249 C

$q=2.49\times 10^{-5}\ C$

Hence, this is the required solution.

5 0

3 years ago

An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the

Lena [83]

Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec = 78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

u=0m/s t=1 s a=9.8m/s²

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

u=0m/s t=2 s a=9.8m/s²

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

u=0m/s t=3 s a=9.8m/s²

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

u=0m/s t=4 s a=9.8m/s²

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

u=0m/s t=5 s a=9.8m/s²

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

7 0

3 years ago

A train accelerates from 23m/s to 190m/s in 54 seconds. What was its acceleration?

Ierofanga [76]

Answer:

Use the method on the image and solve it.

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4 years ago

Cart 1 has an initial velocity and hits cart 2 which is stationary. after a perfectly inelastic collision, the combined carts ar

Tomtit [17]

Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.

The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.

Let's demonstrate the issue using variables:

Let,

m1=mass of cart 1

m2=mass of cart 2

v1 = velocity of cart 1 before collision

v2 = velocity of cart 2 before collision

v' = velocity of the carts after collision

Using the conservation of momentum for perfectly inelastic collisions:

m1v1 + m2v2 = (m1 + m2)v'

v2 = 0 because it is stationary

v' = 1/3*v1

m1v1 = (m1+m2)(1/3)(v1)

m1 = 1/3*m1 + 1/3*m2

1/3*m2 = m1 - 1/3*m1

1/3*m2 = 2/3*m1

m2 = 2m1

From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.

To learn more about inelastic collision visit:

brainly.com/question/14521843

#SPJ4

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2 years ago