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3 years ago

Hypothesis (5 points)

Physics

1 answer:

adelina 88 [10]3 years ago

3 0

Answer:

pretty sure the answer is speed walk, hope this helps

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A space probe lands on a newly discovered planet. A small canister is released from the probe and falls a distance of 3 m in 0.5

ivanzaharov [21]

Answer:

24m/s²

Explanation:

Given

Distance S = 3m

Time of fall = 0.5sec

Required

Acceleration due to gravity

Using the equation of motion

S = ut+1/2gt²

Substitute the given values

3 = 0+1/2g(0.5)²

3 = 1/2(0.25)g

3 = 0.125g

g = 3/0.125

g = 24

Hence the value for the acceleration of gravity on this new planet is 24m/s²

3 0

2 years ago

Help for brainlist easy science plz

zalisa [80]

Answer:

the pics upside down fam

Explanation:

6 0

2 years ago

Read 2 more answers

A boy standing on the bridge kicks a stone into the water below. He kicks the stone with a horizontal velocity of 8.06 m/s. It l

Ronch [10]

Time taken to reach water :

$t = \dfrac{D}{v}\\\\t=\dfrac{6.78}{8.06}\ s\\\\t=0.84\ s$

Now, initial vertical speed , u = 0 m/s.

By equation of motion :

$h = ut +\dfrac{at^2}{2}$

Here, a = g = acceleration due to gravity = 9.8 m/s².

So,

$h = 0(t) +\dfrac{9.8\times 0.84^2}{2}\\\\h=3.46\ m$

Therefore, the height of the bridge is 3.46 m.

Hence, this is the required solution.

6 0

2 years ago

One car travels 56 meters due east in 2.0 seconds, and a second car travels 84 meters due west in 3.0 seconds. During their peri

Lubov Fominskaja [6]

Answer:

average speed

Explanation:

The directions were different, so the velocities could not be the same.

However, the magnitude of the velocity (speed) was 56/2 = 28 m/s for the first car, and 84/3 = 28 m/s for the second car. These<em> average speeds are the same</em>.

5 0

3 years ago

A cannon ball is shot horizontally off a 37.0 m cliff and lands a distance of 18.5 m

4vir4ik [10]

Answer:

vₓ = 6.73 m/s

Explanation:

Assuming no other external influences than gravity, in the horizontal direction (which we make to coincide with the x- axis) , speed is constant, so, applying the definition of average velocity, we can write the following equation:

$v_{x} = \frac{\Delta x}{\Delta t} (1)$

Now, in the vertical direction (coincident with the y- axis) , as both movements are independent each other, initial velocity is zero, so we can write the following equation for the vertical displacement:

$\Delta h = \frac{1}{2} * g * t^{2} (2)$

where Δh = -37.0 m , g = -9.8 m/s2
Solving (2) for t, we get:

$t = \sqrt{\frac{2*\Delta h}{g} } =\sqrt{\frac{2*37.0m}{9.8m/s2}} = 2.75 s (3)$

Taking t₀ = 0, ⇒ Δt = t
Replacing (3) in (1), we get:

$v_{x} = \frac{\Delta x}{\Delta t} = \frac{x}{t} = \frac{18.5m}{2.75s} = 6.73 m/s$

As the horizontal velocity is constant, the initial horizontal velocity is just the average one, i.e., 6.73 m/s.

5 0

2 years ago