Answer:
23 cm
Explanation:
The formula for magnification is;
Magnification = image distance / object distance
use values as;
3.2 = v / 30 where v is image distance
v =30*3.2
v=96 cm
The relationship of the focal length with image distance and object distance is expressed as;

where f is the focal length and u is object distance
use values in the equation as;


f=160/7
f=22.86
f= 23 cm ----------nearest a cm
Answer:
The maximum height is determined by the initial vertical velocity. Since steeper launch angles have a larger vertical velocity component, increasing the launch angle increases the maximum height.
Explanation:
brainliest plzzzzzzzzzzzzzzz
Answer:
Mass = 1350kg
Explanation:
Given the following data;
Velocity = 23m/s
Momentum = 31,050 kg·m/s.
To find the mass of car;
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
Momentum = mass * velocity
Substituting into the equation, we have
31,050 = mass*23
Mass = 31,050/23
Mass = 1350kg
Therefore, the mass of the car is 1350 kilograms.
The radius of the sphere is r=5.15 cm=0.0515 m, and its surface is given by

So the total charge on the surface of the sphere is, using the charge density

:

The electrostatic force between the sphere and the point charge is:

where
ke is the Coulomb's constant
Q is the charge on the sphere

is the point charge
r is their separation
Re-arranging the equation, we can find the separation between the sphere and the point charge:
Answer:
Time is 14.8 s and cannot landing
Explanation:
This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing are the same
Calculate the time to stop, where it has zero speed
Vf² = Vo² + a t
t = - Vo² / a
t = - 110²/(-7.42)
t = 14.8 s
This is the time it takes to stop the jet
Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero
Vf² = Vo² + 2 to X
X = -Vo² / 2 a
X = -110² / 2 (-7.42)
X = 815.4 m
Since this distance is greater than the length of the runway, the jet cannot stop
Let's calculate the speed you should have to stop on a track of this size
Vo² = 2 a X
Vo = √ (2 7.42 800)
Vo = 109 m / s
It is conclusion the jet must lose some speed to land on this track