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Genrish500 [490]

3 years ago

12

A football is kicked from ground level with an initial velocity of 23.6 m/s at angle of 57.5° above the horizontal. How long, in

seconds, is the football in the air before it hits the ground? Ignore air resistance.

1 answer:

aniked [119]3 years ago

3 0

Answer:

6 seconds is the best answer

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Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0

GuDViN [60]

Answer:

20 hertz of frequency produced.

Explanation:

$\boxed{frequency = \frac{1}{period} }$

Here we will find frequency and period should be in second, here given: 0.05 seconds

using the formula:

$\boxed{frequency = \frac{1}{0.05} }$

$\boxed{frequency =20}$

8 0

3 years ago

What is the difference between 1D motion and 2D motion?<br> I need explanation, pleases.

zaharov [31]

Explanation:

1D motion is motion in only one direction. To put it simply, it's motion along a line. For example, a train on a straight track has 1D motion.

2D motion is motion in two directions. A good example of this is a projectile launched at an angle. It has both vertical motion and horizontal motion.

5 0

3 years ago

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Determine the minimum angle at which a roadbedshould be banked

poizon [28]

To solve this problem, apply the concepts related to the relationship given between the centripetal Force and the Weight.

The horizontal force component is equivalent to the weight of the car, while the vertical component is linked to the centripetal force exerted on the car, therefore,

$T cos\theta = mg \rightarrow T = \frac{mg}{cos\theta}$

$Tsin\theta = \frac{mv^2}{r} \rightarrow T = \frac{mv^2/r}{sin\theta}$

Equating both equation we have that,

$\frac{mv^2}{r} = mgtan\theta$

$tan\theta = \frac{v^2}{rg}$

Rearranging to find the angle we have that,

$\theta = tan^{-1} (\frac{v^2}{rg})$

Our values are given as,

$r = 2.00*10^2m$

$v = 20m/s$

$\theta = tan^{-1}(\frac{20^2}{(2*10^{2})(9.8)})$

$\theta = 11.53 \°$

Therefore the minimum angle will be 11.53°

5 0

3 years ago

The contour interval of a topographic map is 25 feet. How many contour lines will be drawn between sea level and an elevation of

Juliette [100K]

<span>If a contour interval of a topographic map is 50 feet, there would be 3 contour lines drawn between sea level and an elevation of 175 feet.</span>

8 0

3 years ago

In the figure a block slides along a track from one level to a higher level, by moving through an

Scrat [10]

Answer:

(B) 1.17 m.

Explanation:

At the highest point, the conservation of energy equation would becomes;

K.E (initial ) = K.E + P.E;

½ m(V)^2 = ½ m(V)^2 + mgh

(m*6^2)/2 = (m * 9.81 * 1.1) + (m*V^2)/2

36/2 = (9.81 * 1.1) + v^2/2

Solving the equation;

v=√14.44 m/s

= 3.8 m/s

a = −0.6g

= -0.6*9.81

= −5.88 m/s^2

Using equation of motion,

V^2 = U^2 + 2aS

S = V^2/2a

= 14.44/12

= 1.19 m.

7 0

4 years ago