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2 years ago

A ball is fired vertically into the air. Its height in feet t seconds later is given by the formula: h(t)= -2t^2+80t

1 answer:

4 0

v(t)=40-16t 
so max height is when v(t)=0 or t=40/16 
and the max height is found by plugging that into the height equation

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An 8.50 m long ladder leans against the side of a building. The ladder is initially inclined at an angle of 47.0° to the horizon

erastova [34]

The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°

cos θ = Adjacent side / Hypotenuse

θ $_{1}$ = 47°

Hypotenuse = Length of ladder = 8.5 m

cos 47° = Adjacent side / 8.5

Adjacent side = Initial distance of base of ladder from the building = 5.8 m

Adjacent side 2 = Final distance of base of ladder from the building

Adjacent side 2 = 5.8 - 0.82 = 4.98 m

cos θ $_{2}$ = Adjacent side 2 / Hypotenuse

cos θ $_{2}$ = 4.98 / 8.5 = 0.59

θ $_{2}$ = $cos^{-1}$ ( 0.59 )

θ $_{2}$ = 53.84°

The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.

Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°

To know more about trigonometric ratios

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3 0

2 months ago

What occurs when a swimmer pushes through water to swim

Triss [41]

When we swim we apply force and push the water backward with the help of our hands. In response, The water pushes us forward with an equal force. Thus, in order to move forward and swim, the swimmer lushes the water backward. Newton's 3rd law of motion

6 0

2 years ago

A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the

Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0

2 years ago

A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving w

Nadusha1986 [10]

C an s and yes ihavetotypemore

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2 years ago

Find the velocity of a wave that has a wavelength of 2 meters and a frequency of 2.1Hz

vodomira [7]

It is 4.2 hope i helped

7 0

2 years ago