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2 years ago

Answer fast , is si unit newton/meter square or pascal or both

Physics

1 answer:

Readme [11.4K]2 years ago

8 0

Answer:

Both

Explanation:

The S.I. unit of pressure is newton/meter square or pascal as both represent the same dimensional value.

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A football is kicked from ground level at an angle of 53 degrees. It reaches a maximum height of 7.8 meters before returning to

Nonamiya [84]

Answer:

1.61 second

Explanation:

Angle of projection, θ = 53°

maximum height, H = 7.8 m

Let T be the time taken by the ball to travel into air. It is called time of flight.

Let u be the velocity of projection.

The formula for maximum height is given by

$H = \frac{u^{2}Sin^{2}\theta }{2g}$

By substituting the values, we get

$7.8= \frac{u^{2}Sin^{2}53 }{2\times 9.8}$

u = 9.88 m/s

Use the formula for time of flight

$T = \frac{2uSin\theta }{g}$

$T = \frac{2\times 9.88\times Sin53 }{9.8}$

T = 1.61 second

7 0

2 years ago

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

<u>Where</u>:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0

3 years ago

What allows a metal to be pounded into a flat shape?

vovikov84 [41]

C im pretty sure?.........

3 0

3 years ago

A 35-year-old patent clerk needs glasses of 50-cm focal length to read patent applications that he holds 25 cm from his eyes. Fi

Natali5045456 [20]

Answer:

28.57 cm

Explanation:

We are given that

Focal length, $f=50 cm$

Distance of application from his eyes,s=40 cm

$\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}$

$\frac{1}{40}+\frac{1}{s'}=\frac{1}{50}$

$\frac{1}{s'}=\frac{1}{50}-\frac{1}{40}=\frac{4-5}{200}$

$s'=\frac{200}{-1}=-200 cm$

s=25 cm

Substitute the values

$\frac{1}{25}-\frac{1}{200}=\frac{1}{f'}$

$\frac{8-1}{200}=\frac{1}{f'}$

$\frac{1}{f'}=\frac{7}{200}$

$f'=\frac{200}{7}=28.57 cm$

6 0

3 years ago

Identical cannonballs are fired with the same force, one each from four cannons having respective bore lengths of 1.0 meter, 2.0

ANEK [815]

I think its options 2 idk man

6 0

3 years ago

Read 2 more answers