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Scrat [10]
3 years ago
15

A fluid is any substance that can flow and take the shape of its container A) true B) false

Physics
2 answers:
Tpy6a [65]3 years ago
7 0
Its true...water takes the shape of its container and it flows..please give me 5 stars and brainliest

Natalka [10]3 years ago
3 0

it is A) True

Hoped I helped


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A ball on the end of a string is whirled around in a horizontal circle of radius 0.478 m. The plane of the circle is 1.02 m abov
m_a_m_a [10]

Answer:

Explanation:

We shall find first the velocity of ball at the time when string breaks. Let it be v . During its fall on the ground , 1.02 m below, we use the formula

h = 1/2 gt² where t is time of fall .

1.02 = 1/2 x 9.8 x t²

t²= .2081

t = .456

During this time it travels horizontally at distance of 2.5 m with uniform velocity of v

v x .456 = 2.5

v  = 5.48 m /s

centripetal acceleration

= v² / r where r is radius of the circular path

= 5.48² / .478

= 62.82 m /s²

3 0
3 years ago
What is the difference between muscular strength and muscular endurance
MAXImum [283]
Muscular strength is a measure of how much force you can exert in one repetition. Muscular endurance refers to the ability to perform a specific muscular action for a prolonged period of time.
4 0
3 years ago
Read 2 more answers
Why is it important we uncover and study fossils?
Ilia_Sergeevich [38]

its important so we can learn things about the species  

4 0
3 years ago
Read 2 more answers
Can someone please help
ki77a [65]

Answer:

Acceleration of that planet is 30 \frac{m}{s^{2} }.

Given:

initial speed of hammer = 0 \frac{m}{s}

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + \frac{1}{2} g 1^{2}

g = 15 × 2

g = 30 \frac{m}{s^{2} }

Thus, acceleration of that planet is 30 \frac{m}{s^{2} }.

8 0
3 years ago
A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began
Montano1993 [528]

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 =  36.95 m/s

8 0
3 years ago
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