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2 years ago

From the anatomical position, the digits are ____ to the brachium.

Physics

2 answers:

dedylja [7]2 years ago

7 0

I think is a if I’m wrong tell em I’ll re slove this

PIT_PIT [208]2 years ago

4 0

A I believe if not I’m sorry

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Name the type of reproduction process as shown in Fig 1 and Fig 2. State one point of difference between the two

likoan [24]

Figure 1= binary fission in amoeba

figure 2= budding in yeast

difference

1.Parent divides to form two daughter cells and itself gets disappeared in binary fission but in budding , a bud gets matured and detaches from the parent

6 0

3 years ago

Rahim is watching his favorite football team on television. In order to work, the television must be plugged into an electrical

Rasek [7]

I think the answer is B

8 0

3 years ago

Read 2 more answers

To win the game, a place kicker must kick a

Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance. Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement. Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0

3 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$