Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.
Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:
1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.
replacing the values we obtain:
1/8.3+1/q=1/15.2
so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3
then q=-18.28 cm
The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2
We also add a picture to see the imagen formation for this case.
Speed of man towards West =2.5 m/s
Time to travel in the given direction = 1200 s
Total displacement towards west
Now speed of man towards East = 1 m/s
time to travel in east direction = 500 s
total displacement towards East
Now total displacement will be
so it is 2500 m towards west
Answer:
The ball travels <u>1.31 m horizontally</u> before hitting the ground.
Explanation:
Given:
Initial velocity of the ball is,
Angle of projection is,
Initial height of the ball is,
Final height of the ball is,
Now, horizontal and vertical components of initial velocity are given as:
Plug in the given values and find . This gives,
Now, there is acceleration due to gravity acting in the vertical direction. The direction is downward. So, .
Now, using equation of motion in the vertical direction, we have:
Plug in the given values and solve for time 't'. This gives,
On solving the above quadratic equation, we get:
Ignoring the negative result as time can't be negative. So, time taken by the ball to reach the ground is 0.57 s.
Now, as the ball moves, there is no force acting on it in the horizontal direction. So, the acceleration in the horizontal direction is 0.
Now, we know that, when acceleration is zero, the body moves with a constant speed.
Hence, distance traveled in the horizontal direction is given as:
Horizontal distance = Horizontal component of initial velocity × Time
Plug in the given values and solve for 'R'. This gives,
So, the ball travels 1.31 m horizontally before hitting the ground.
Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 × m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 = Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 = + 1000 × 9.8 ×36
pressure at bottom P2 =4.52 × Pa
so from gas law
here p is pressure and v is volume and t is temperature
so put here value and find v1
V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³