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Jobisdone [24]
3 years ago
14

A ball is thrown from 1 m above the ground. The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal. What

is the maximum height of the ball above the ground
Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

9.4 m

Explanation:

This is a projectile motion. The maximum height above the level projection is given by

H = \dfrac{(v_0\sin\theta)^2}{2g}

v_0 is the initial velocity, \theta is the angle of projection above the horizontal and g is the acceleration of gravity.

H = \dfrac{(20\sin40^\circ)^2}{2\times9.8} = \dfrac{165.27}{19.6} = 8.4\text{ m}

The maximujm height above the ground is 1 m + 8.4 m = 9.4 m

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artcher [175]
<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            -------------------(i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               -----------------------(ii)

<em>But;</em>

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

<em>From equation (ii), it implies that;</em>

F = W = m x g

<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e      ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

k = 278N/m

<em>Substitute these values into equation (iii) as follows;</em>

3.5 x 10 = 278 x e

35 = 278e

<em>Now solve for e;</em>

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

3 0
4 years ago
Assume that the polymer material has a constant refractive index of 1.5. For light of 600nm wavelength at normal incidence, what
yaroslaw [1]

Answer:

Minimum thickness will be 100 nm

Explanation:

We have given refractive index is n = 1.5

Wavelength of the light incidence \lambda= 600 nm

We have to find the smallest thickness of the film so that there will be minimum light reflect

For minimum thickness of non reflecting film

t=\frac{\lambda }{4n} , here t is thickness, \lambda is wavelength and n is refractive index

Putting all values t=\frac{600}{4\times 1.5}=100nm

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Answer:

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