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3 years ago

Give an example of the follow components of culture in your life and explain the role it plays.

Physics

1 answer:

zhenek [66]3 years ago

7 0

Symbols such as cross motivates me to go to church.

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If a ball is thrown into the air with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. fi

Juliette [100K]

Height (y) = 36t - 16t^2, where t = time in seconds (s).
Our height (y) after 1s = 36(1) - 16(1)^2
y = 36 - 16 = 20 ft
So it reached a height of 20 ft during that 1 second, which means that at that 1 second it had a velocity of 20ft/s, since v = d(distance)/t = 20ft/1s

6 0

3 years ago

In Newton’s second law, if the net force acting on object doubles. The object’s Will also double

Anna71 [15]

Answer: Explanation:

If the net force on an object is doubled, its acceleration will double If the mass of an object is doubled, the acceleration will be halved .

3 0

3 years ago

An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it

KATRIN_1 [288]

Answer:

v = 7.4 m/s

Explanation:

Given that,

Mass if a volleyball, m = 5 kg

The ball reaches a height of 2.8 m

We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 2.8} \\\\=7.4\ m/s$

So, the required speed is 7.4 m/s. Hence, the correct option is (b).

6 0

3 years ago

- How much force is needed to accelerate a 26 kg skier at 4 m/sec??

kondor19780726 [428]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 26 × 4

We have the final answer as

Hope this helps you

8 0

3 years ago

a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg

IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

$M = \frac{q}{p}$

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

$4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m$

Now using thin lens formula:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\$

6 0

3 years ago