Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )
x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)

Answers

Answer 1

Answer:

The formula for calculating the horizontal displacement of a horizontally launched projectile is $x=v*t$

A projectile launched horizontally with a velocity v, at a height y ,travels a horizontal distance x, while falling through a distance y. The horizontal velocity of a projectile remains constant throughout its motion, in the absence of air resistance. The vertical component of the velocity is under the action of the gravitational force and hence it increases in magnitude as it falls through the height.

The horizontal displacement of the projectile is a uniform motion and it occurs at a constant speed v.

Thus, the horizontal displacement of the projectile is given by the expression.

$x=v*t$

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Power P is the rate at which energy E is consumed per unit time. Ornithologists have found that the power consumed by a certain pigeon flying at velocity v m/s is described well by the function P(v)=16v−1+10−3v3 J/s. Assume that the pigeon can store 5×104 J of usable energy as body fat. Find the velocity vp min that minimizes power consumption. (Use decimal notation. Give your answer to two decimal places.)

Answers

Answer:

Explanation:

P(v) = 16 / v + 10⁻³ v³

differentiating on both sides

dP / dt = - 16 / v² + 3 x 10⁻³ v²

For maxima and minima , the condition is

dP / dt = - 16 / v² + 3 x 10⁻³ v² = 0

v² = 160 / 3 x 10²

v² = 73 m/s

v = 8.54 m /s

To know the condition of minima

again differentiating

d²P / dt² = - 16 x -2 / v² + 6 x 10⁻³ x v

= 32 / v³ + 6 x 10⁻³ x v

= + ve quantity

So at v_p = 8.54 m /s , power consumption will be minimum .

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle with a constant angular speed of 120 rev/min. The cross-sectional area of thewire is 0.014 cm2. Calculate the elongation of the wire when the mass is a) at its lowest point of the path and b) at the highest point of its path

Answers

Answer:

a) the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

b) the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

Explanation:

Given that;

the angular speed $\omega = 120 \ rev/min$

Then converting it to rad/s ; we have:

= $(120 \ rev/min )((2 \ \pi \ rad )/(1 \ rev) ) ((1 \ min )/(60 \ s) )$

= 12.57 rad/s

The cross-sectional area of the wire A = 0.014 cm²

A = (0.014 cm²) ( $(10^(-4) \ m^2)/(1 \ cm^3)$ )

A = $0.014*10^(-4) \ m^2$

mass (m) = 12.0 kg

R = 0.5 m

g = 9.8 m/s²

To calculate for the mass when its at the lowest point of the path; we use the Newton's second law of motion; which is expressed as:

$T - mg = ma_(rad)$

where;

$a_rad = ( radical \ acceleration ) = R \omega^2$

Now; we can rewrite our equation as;

$T -mg = m R \omega ^2$

$T = mR \omega^2 + mg$

$T = m( R \omega^2 + g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 + 9.8)$

$T = 12.0( 0.5(158.0049) + 9.8)$

$T = 12.0( 79.00245 + 9.8)$

$T = 12.0( 88.80245)$

T = 1065.6294 N

T ≅ 1066 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((1066 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00544 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.5 cm

Thus, the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

Using Newton's second law of motion also for the mass at its highest point of the path; we have:

$T +mg = m R \omega ^2$

$T = mR \omega^2- mg$

$T = m( R \omega^2 - g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 - 9.8)$

$T = 12.0( 0.5(158.0049)-9.8)$

$T = 12.0( 79.00245 - 9.8)$

$T = 12.0( 69.20245)$

T = 830.4294 N

T = 830 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((830 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00424 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.42 cm

Thus, the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

A positive magnification means the image is inverted compared to the object

Answers

False

Explanation:

A positive magnification means the image is erect compared to the object. Magnifications with values greater than one represent images that are smaller than the object. A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate the power absorbed by any surface. The equation applies only to perfect radiators. The equation applies only to perfect absorbers. The equation is valid with any temperature units. The equation describes the transport of thermal energy by conduction.

Answers

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

Explanation:

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.

$P=e \sigma A\left(T^(4)-T_(c)^(4)\right)$

So this law is application for calculating power absorbed by any surface.

A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m

Answers

Answer:

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Explanation:

Considering vertical motion of cat:-

Initial velocity, u = 3.44 sin60 = 2.98 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v² = u² + 2as

Substituting

v² = u² + 2as

0² = 2.98² + 2 x -9.81 x s

s = 0.45 m

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Go to his profile and roast the mess out of him plzz 403665fl 50 points

Answers

Answer:

Explanation:

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Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )x = vi(cos )tx = aytx = vxt (RIGHT ANSWER)

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Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )
x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)