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juin [17]
3 years ago
13

In which situations is heat flow present? Check all that apply.

Physics
2 answers:
Ber [7]3 years ago
8 0

Answer:

A. A substance feels warm.

D. One substance transfers some of its thermal energy to another substance.

E. Two substances differ in temperature.

Explanation:

babunello [35]3 years ago
7 0
A substance feels warm
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1. How far away must you be from a 675 kHz radio station with power 50.0 kW for there to be only one photon per second per squar
goldenfox [79]

a) 0.321 ly

b) 0.321 light years is not far in astronomical terms. Alien life would need to transmit at tremendous power in order for their radio transmissions to be detectable. Their radio signal also needs to be stronger than background noise in order to be distinguishable. Therefore it is unlikely that radio transmissions from alien life will ever be detected.

6 0
2 years ago
A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0
3 years ago
A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. the clam lands on the roc
vovangra [49]

Part a)

in horizontal direction there is no gravity or no other acceleration

so in horizontal direction the speed of clam will remain same

v_x = 2.70 m/s

Part b)

In vertical direction we can use kinematics

v_f = v_i + at

v_f = 0 + 2.1 * 9.8

v_f = 20.6 m/s

part c)

if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed

5 0
3 years ago
An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 14.0 m if her initial speed is 2.20 m/
pochemuha

We have the equation of motion

                          v^2 = u^2+2as , where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

In this case initial velocity = 2.20 m/s, final velocity = 0 m/s, displacement = 14 m

On substitution we will get 0 = 2.20^2+2*a*14

On solving we can find the acceleration value as -0.173m/s^2

So free fall acceleration = 0.173m/s^2

                                             


3 0
3 years ago
When liquid water gets into cracks of rock and freezes, it __ and ___.
hoa [83]
It expands and pushes the crack further aprt
3 0
2 years ago
Read 2 more answers
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