How many seconds are there in 3 days 38 hrs? Explain.

39.Explain the working mechanism of a mercury barometer.

alina1380 [7]

One end is sealed to prevent air from disturbing the scale and measures. Therefore, a Mercury Barometer works on the principle of balancing the atmospheric pressure with the volume of mercury present in the device

7 0

3 years ago

A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally awa

Ivanshal [37]

Answer:

V=34.2 m/s

Explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :

We know that

$V^2_y=U^2_y+2 g h$

Now by putting the values in the above equation we got

$V^2_y=U^2_y+2 g h$

$V^2_y=0^2+2\times 9.81 \times 54$

Assume $g= 9.81 m/s^2$

Thus

$V^2_y=1059.48$

$V_y=\sqrt{1059.48}\ m/s$

$V_y=32.54 m/s$

We also know that

$V_y=U_y+ g\times t$

$32.54=9.81\times t$

$t=\dfrac{32.54}{9.81}=3.31\ s$

In horizontal direction :

$x=U_x\times t$

$U_x=\dfrac{35}{3.31}=10.54\ m/s$

Thus the resultant velocity

$V=\sqrt{V^2_y+U^2_x}$

$V=\sqrt{32.54^2+10.54^2}=34.2\ m/s$

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.

5 0

4 years ago

4. The figure below shows a complex wave pattern on a string

Marat540 [252]

Explanation:Consider the wave pattern image reflected about the rigid hook on the wall. · 005(part2of3)10.0 pointsHow many complete waves are emitted in this time

5 0

3 years ago

Puck 1 is moving 10 m/s to the left and puck 2 is moving 8 m/s to the right. They have the same mass, m.

Julli [10]

Answer:

(a) the total momentum of the system before the collision = -2m kg.m/s.

(b) the total momentum of the system after the collision = -2m kg.m/s.

(c) puck 1's velocity after the collision in component form = (5.44 i, 2.54 j)

Explanation:

Given;

mass of Puck 1 , = m

mass of Puck 2, = m (since they have the same mass m)

initial velocity of Puck 1, u₁ = 10 m/s to the left

initial velocity of Puck 2, u₂ = 8 m/s to the right

Let the rightward direction be positive direction

Let the leftward direction be negative direction

(a) the total momentum of the system before the collision;

P₁ = (initial momentum of Pluck 1) + (initial momentum of Pluck 2)

P₁ = (-mu₁) + mu₂

P₁ = mu₂ - mu₁

P₁ = m(u₂ - u₁)

P₁ = m(8 - 10)

P₁ = -2m kg.m/s

(b) the total momentum of the system after the collision;

Based on the principle of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision.

Thus, the total momentum of the system after the collision is -2m kg.m/s.

(c) puck 1's velocity after the collision in component form

$v = (v_x, v_y)\\\\v = (vcos \theta , vsin \theta)\\\\v = (6cos 25^0 , 6sin25^0)\\\\v = (5.44i, 2.54j)m/s$

8 0

3 years ago

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

gizmo_the_mogwai [7]

<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>

5 0

3 years ago