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Karo-lina-s [1.5K]

3 years ago

15

Newtons second law states that Force=Mass × Acceleration.The acceleration of gravity in 9.8 m/s.How much force is gravity exerti

ng on the 100g ball

1 answer:

Naddika [18.5K]3 years ago

5 0

.98 Newton’s because you convert 100 g to kg which is .1 kg them you multiply.1 kg by 9.8 and get .98 and the units of the force are in Newton’s

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An echo is an example of sound wave A) diffraction. B) interference. C) reflection. D) refraction.

DedPeter [7]

It is an example of reflection of wave. C.

6 0

3 years ago

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A marble column of cross-sectional area 1.6 m^2 supports a mass of 26600 kg. The elastic modulus for marble is 5.0 times 10^10 N

vodka [1.7K]

Answer:

Δ L = 2.57 x 10⁻⁵ m

Explanation:

given,

cross sectional area = 1.6 m²

Mass of column = 26600 Kg

Elastic modulus, E = 5 x 10¹⁰ N/m²

height = 7.9 m

Weight of the column = 26600 x 9.8

= 260680 N

we know,

Young's modulus= $\dfrac{stress}{strain}$

stress = $\dfrac{P}{A}$

= $\dfrac{260680}{1.6}$

= 162925

strain = $\dfrac{\Delta L}{L}$

now,

$Y = \dfrac{stress}{strain}$

$\Delta L = \dfrac{162925}{Y}\times L$

$\Delta L = \dfrac{162925}{5 \times 10^10}\times 7.9$

Δ L = 2.57 x 10⁻⁵ m

The column is shortened by Δ L = 2.57 x 10⁻⁵ m

3 0

2 years ago

A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at

Rasek [7]

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

$W_{all}=Change\ in\ kinetic\ energy\ of\ system$

We know that kinetic energy of moving mass given as

$KE=\dfrac{1}{2}mv^2$

Now by pitting the values

$KE=\dfrac{1}{2}mv^2$

$KE=\dfrac{1}{2}\times 0.011\times 29^2$

KE= 4.25 J

So the work done on the ball is 4.25 J

8 0

3 years ago

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th

valina [46]

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started. So the height of the cliff is 80 m.

5 0

2 years ago

a0.155 kg arrow is shot from ground level, upward at 31.4 m/s. what is its kinetic energy (ke) when it is 30.0 m above the groun

swat32

The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J

Assuming air friction is negligible,

a = - 9.8 m / s²

u = 31.4 m / s

s = 30 m

v² = u² + 2 a s

v² = 31.4² + ( 2 * - 9.8 * 30 )

v² = 985.96 - 588

v² = 397.96 m / s

KE = 1 / 2 m v²

KE = 1 / 2 * 0.155 * 397.96

KE = 0.0775 * 397.96

KE = 30.85 J

Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J

To know more about kinetic energy

brainly.com/question/24360064

#SPJ1

5 0

1 year ago