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velikii [3]
3 years ago
10

3. An element that is malleable is

Physics
2 answers:
user100 [1]3 years ago
7 0

Answer:

Difficult to Pour.

Explanation:

lbvjy [14]3 years ago
6 0
It is A. Easily shaped
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A 2kg bowling ball is 2.5 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7m/s.
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3 years ago
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Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0
3 years ago
Can someone help with this???
Rus_ich [418]

no BECQUSE POSUM BROOB SHSHSJ

8 0
2 years ago
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B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

V_{int}=\sqrt{U^2+19.6H}

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration  is mathematically given as

a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2

Where

acceleration is still -9.81 m/s2,

Hence,

V^2-U^2=2(-9.81)*-H

Therefore

V_{int}=\sqrt{U^2+19.6H}

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

7 0
2 years ago
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