True or false cold air can hold more moisture than warm air

At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another b

maks197457 [2]

Answer:

(a) The two balls collide $2\; \rm s$ after launch.

(b) The height of the collision is $4\; \rm m$ .

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of $g$ should be negative. The question states that the magnitude of $g\!$ is $10\; \rm m \cdot s^{-2}$ . Hence, the signed value of $\! g$ should be $\left(-10\; \rm m \cdot s^{-2}\right)$ .

Similarly, the initial velocity of the ball thrown downwards should also be negative: $\left(-8.0\; \rm m \cdot s^{-1}\right)$ .

On the other hand, the initial velocity of the ball thrown upwards should be positive: $\left(12\; \rm m \cdot s^{-1}\right)$ .

Let $v_0$ and $h_0$ denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time $t$ :

$\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0$ .

For both balls, $g = \left(-10\; \rm m \cdot s^{-2}\right)$ .

For the ball thrown downwards:

Initial velocity: $v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)$ .
Initial height: $h_0 = 40\; \rm m$ .

$\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40$ (where $h$ is in meters and $t$ is in seconds.)

Similarly, for the ball thrown upwards:

Initial velocity: $v_0 = \left(12\; \rm m \cdot s^{-1}\right)$ .
Initial height: $h_0 = 0\; \rm m$ .

$\displaystyle h(t) = -5\, t^{2} + 12\, t$ (where $h$ is in meters and $t$ is in seconds.)

Equate the two expressions and solve for $t$ :

$-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t$ .

$t = 2$ .

Therefore, the collision takes place $2\, \rm s$ after launch.

Substitute $t = 2$ into either of the two original expressions to find the height of collision:

$h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m$ .

In other words, the two balls collide when their height was $4\; \rm m$ .

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3 years ago

Angular oscillation of the slotted link is achieved by the crank OA, which rotates clockwise at the steady speed N = 118 rev/min

Leni [432]

Answer:

Angular velocity will be 65.3 rad/s

7 0

3 years ago

Read 2 more answers

In an experiment of a simple pendulum, measurements show that the pendulum has length し 0.397 ± 0.006 m, mass M-0.3172 ± 0.0002

vichka [17]

Answer:

1.)1.265+or minus 0.0006m

2).0.71%

Explanation:

See attached file

6 0

3 years ago

Give three examples that show the importance of gravity to humans

kow [346]

1-Without gravity you can't grow any plant or tree because dirt would fly away
2-We are using gravity for a lot of physics system too
3-All the water on Earth would have been flying in the space
4-There wouldn't be atmosphere.

3 0

3 years ago

the potential energy of a 40kg cannon ball is 14000j. How high was the cannon ball to have this much potential energy?

noname [10]

We will use the formula p = mgh p is potential energy. m is mass of object in kg g is acceleration due to gravity (9.8m/s²) h is height of the objects displacement in meters. p = mgh → mgh = p → h = p / mg p is 14000j, m is 40kg and g is 9.8 m/s² h = 14000 / 40 × 9.8 → h = 1400 / 392 → h = 35.7 Therefore , the cannonball was 35.7 meters high .

7 0

3 years ago