A box contains about 6.77 x 1021 hydrogen

A student conducted an experiment that used two rose plants to determine if a liquid formula for fertilizer was better than a dr

Serga [27]

B. it was necessary to see data from more plants before the conclusion could be accepted.

4 0

3 years ago

Dans un tube en U contenant du mercure ,on verse de l'autre côté de l'acide sulfurique de densité 1,84 et de l'autre côté de l'a

Iteru [2.4K]

Explanation:

Unclear question. The clear rendering reads;

"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?

6 0

3 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

6 0

3 years ago

Pls help I need this done soon 50 points

lidiya [134]

Answer:

1,3,5

Explanation:

i think maybe dont come at me

6 0

3 years ago

Read 2 more answers