PHYSICS COLLEGE

A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the projectile a freely falling body?Yes or No(b) What is its acceleration in the vertical direction? (Let up be the positive direction.)____? m/s2(c) What is its acceleration in the horizontal direction?

Answers

Answer 1

Answer:

Answer:

A) No

B)-9,81 m/s^2

C)0 m/s^2

Explanation:

A free fallin object has only velocity on the vertical axis so any object that is moving in the Y and X axis has projetile motion not free falling, and when dealing with projectile motion the object is experiencing acceleration towards the ground of -9,81m/s^2 and in the Y axis, in the X axis there´s is only acceleration if the air is providing resistance, since it states that it isnot, then the accleration is 0.

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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 .

1 microgram equals how many milligrams?

Answers

Answer: 1 microgram is equal to 0.001 miligrams

Explanation: The factor micro is equal 10^-3 while the factor mili is equal to 10^-3 so to converte the micro to mile we have to multiply by 0.001.

The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)

Answers

This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

$W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)$

Taking the derivative with respect to actual temperature, we get:

$(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n$

where,

dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h

Therefore,

$dW=(-1)(0.6215-0.3965(18)^(0.16))$

dW = - 1.25°C

Learn more about derivatives here:

brainly.com/question/9964510?referrer=searchResults

Answer:

Δw=1.25°C

Explanation:

Given that

$w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)$

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

$(dw)/(dT)=0.6215 +0.3965v^(0.16)$

$(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)$

Now by putting the values v=19 km/hr and ΔT=1

$(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)$

Δw=1.25°C

So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:
d1=_____m
Part B:
d2=______m

Answers

Answer:

Explanation:

In projectile motion , range of projectile is given by the expressions

R = u²sin2θ / g

where u is velocity of projectile.

u = 27 m/s θ = 50

12 = 27² sin 2θ / 9.8

sin 2θ = .16

θ = 9.2 / 2

= 4.6

When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.

HELP ASAP PLEASE!!!In which direction(s) does the ground shake during an earthquake?
A. sideways
B. up and down
C. back and forth
D. all of the above

Answers

Answer: D i am pretty sure

Explanation:

Answer:

all

Explanation:

Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off