How do smaller summer ice flores affect polar bears physically?

A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of

Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

$\eta = 1-\frac{T_L}{T_H}$

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

$\eta = 1-\frac{T_L}{T_H}$

$\eta = 1-\frac{293}{713}$

$\eta = 0.589$

Therefore the maximum possible efficiency the car can have is 58.9%

4 0

3 years ago

17. A 1350 g projectile is launched with a force of 150 N. The length of the firing arm is 1.25 m.

creativ13 [48]

a) The exit velocity of the projectile is 16.7 m/s

b) The maximum height achieved by the projectile is 14.2 m

c) The total time of flight is 3.40 s

d) The distance covered by the projectile is 46.5 m

Explanation:

a)

We solve this first part of the problem by applying the work-energy theorem, which states that the work done on the projectile is equal to the gain in kinetic energy of the projectile. Mathematically:

$W=Fd = \frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K$

where:

F = 150 N is the force applied

d = 1.25 m is the displacement of the projectile (the length of the firing arm)

m = 1350 g = 1.35 kg is the mass of the projectile

u = 0 is the initial velocity of the projectile

v is the exit velocity of the projectile

Solving for v, we find:

$v=\sqrt{\frac{2Fd}{m}}=\sqrt{\frac{2(150)(1.25)}{1.35}}=16.7 m/s$

b)

Assuming the projectile is fired vertically upward, then the initial kinetic energy of the projectile as soon as he leaves the cannon is fully converted into gravitational potential energy as it reaches the top of its trajectory. So we can write:

$K_i = U_f$

$\frac{1}{2}mv^2=mgh$

where:

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

m = 1350 g = 1.35 kg is the mass of the projectile

v = 16.7 m/s is the velocity at which the projectile leaves the cannon

$g=9.8 m/s^2$ is the acceleration of gravity

h is the maximum height reached by the projectile

And solving for h, we find

$h=\frac{v^2}{2g}=\frac{(16.7)^2}{2(9.8)}=14.2 m$

c)

Assuming the projectile is launched vertically upward, then the total time of flight is twice the time it takes for reaching the maximum height. This time can be found by using the following suvat equation:

$v=u-gt$

where:

u = 16.7 m/s is the initial velocity

$g=9.8 m/s^2$ is the acceleration of gravity

t is the time

The projectile reaches the maximum height when the vertical velocity becomes zero, so when v = 0. Therefore, substituting,

$0=u-gt\\t=\frac{u}{g}=\frac{16.7}{9.8}=1.70 s$

So, the total time of flight is

$T=2t=2(1.70)=3.40 s$

d)

The motion of a projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

First we have to analyze the vertical motion, to find the time of flight of the apple. We can do it by using the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s = -27 m is the vertical displacement of the apple

$u_y=u sin \theta = (16.7)(sin 42^{\circ})=11.2 m/s$ is the initial vertical velocity

t is the time if flight

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting we have:

$-27=11.2t - 4.9t^2\\4.9t^2-11.2t+27=0$

which has two solutions:

t = -1.46 s (negative, we discarde)

t = 3.75 s (this is our solution)

Now we can analyze the horizontal motion: the projectile moves horizontally with a constant velocity of

$v_x = u cos \theta = (16.7)(cos 42^{\circ})=12.4 m/s$

So, the distance it covers during its fall is given by

$d=v_x t=(12.4)(3.75)=46.5 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

4 years ago

Define longitudinal wave

Alja [10]

longitudinal wave, wave consisting of a periodic disturbance or vibration that takes place in the same direction as the advance of the wave. ... Sound moving through air also compresses and rarefies the gas in the direction of travel of the sound wave as they vibrate back and forth.

<u>FORMULA</u>

y = displacement of the point on the traveling sound wave

x = distance the point has traveled from the wave's source

y_0 = amplitude of the oscillations

\omega = angular frequency of the wave

t = time elapsed

c = speed of the wave

6 0

3 years ago

Explain how you would find the units for acceleration in a problem?

deff fn [24]

Answer:

Calculating acceleration involves dividing velocity by time — or in terms of SI units, dividing the meter per second [m/s] by the second [s]. Dividing distance by time twice is the same as dividing distance by the square of time. Thus the SI unit of acceleration is the meter per second squared .

6 0

3 years ago

In a game of darts, the dart thrower releases the dart directly at the center of the target (the "bull’s eye") with a speed of 1

Nuetrik [128]

Answer:2.81 m

Explanation:

Given

speed(u)=13 m/s

dart strikes 0.23 m below

i.e. vertical distance traveled by dart is 0.23 m

considering dart motion similar to Projectile

$h=ut+\frac{at^2}{2}$