A plane mirror and a concave mirror (f = 8.40cm) are facing each other and are separated by a distance of 22.0 cm. An object is
1 answer:
Answer:
The distance of the image produced by the concave mirror is 11.27 cm.
Explanation:
Given that,
Focal length = 8.40 cm
Distance between object and plane mirror = 11.0 cm
Distance between plane mirror and concave mirror = 22.0 cm
We need to calculate the object distance
We need to calculate the distance of the image
Using formula of distance of the image
Where, u = Object distance
f = focal length
Put the value into the formula
Hence, The distance of the image produced by the concave mirror is 11.27 cm.
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Answer:
Their final relative velocity is 0.190 m/s
Explanation:
The relative velocity of the satellites, v = 0.190 m/s
The mass of the first satellite, m₁ = 4.00 × 10³ kg
The mass of the second satellite, m₂ = 7.50 × 10³ kg
Given that the satellites have elastic collision, we have;
Given that the initial velocities are equal in magnitude, we have;
u₁ = u₂ = v/2
u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s
v₁ and v₂ = The final velocities of the satellites
We get;
The final relative velocity of the satellite, = v₁ + v₂
∴ = 0.095 + 0.095 = 0.190
The final relative velocity of the satellite, = 0.190 m/s
<h2>Hello!</h2>
The answer is:
The kinetic energy of the object is equal to 85 J.
<h2>Why?</h2>The kinetic energy involves the speed and the mass of an object in motion. We can calculate the following the work needed to speed an object (kinetic energy) using the equation:
Where,
m, is the mas of the object
v, is the speed of the object.
Now, we are given:
So, substituting and calculating the kinetic energy of the object, we have:
We have that the kinetic energy of the object is equal to 85 J.
Have a nice day!
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
