the focal length is much more decent for a concave, and also worse for a convex mirror. When the image that is given, distance is good and decent, images are always on the same area of the mirror as the object given , and it is not fake. images distance is never positive , the image is on the oppisite side of the mirror, so the image must be virtual.

7 0

3 years ago

As a science fair project, you want to launch an 800 g model rocket st raight up and hit a horizontally moving target as it pass

fredd [130]

Sum the forces in the y (upward) direction

$\sum F_y = ma$

$F_t +F_w = ma$

$15N - (0.800kg)(9.81m/s^2) = 0.800kg * a$

$a = 8.94 m/s^2$

Applying the kinematic equations of linear motion we have that the displacement as a function of the initial speed, acceleration and time is

$s = v_0t + \frac{1}{2} at^2$

$30 = 0 +\frac{1}{2} (8.94) t^2$

$t = 2.59 s$

Again through the kinematic equation of linear motion that describes velocity as the change of displacement in a given time, we have to

$v = \frac{d}{t} \rightarrow d = vt$

$d = (15m/s)(2.59s)$

$d = 38.85m$

Therefore the horizontal distance between the target and the rocket should be 38.83m

4 0

3 years ago

A cardinal (Richmondena cardinalis) of mass 4.20×10−2 kg and a baseball of mass 0.146 kg have the same kinetic energy. What is t

bekas [8.4K]

Explanation:

It is given that,

Mass of cardinal, $m_c=4.2\times 10^{-2}\ kg$

Mass of baseball, $m_b=0.146\ kg$

Both cardinal and baseball have same kinetic energy. We need to find the ratio of the cardinal's magnitude $p_c$ of momentum to the magnitude $p_c$ of the baseball's momentum.

$E_c=E_b$

Kinetic energy is given by, $E=\dfrac{p^2}{2m}$

$\dfrac{p_c^2}{2m_c}=\dfrac{p_b^2}{2m_b}$

$(\dfrac{p_c}{p_b})^2=\dfrac{m_c}{m_b}$

$\dfrac{p_c}{p_b}=\sqrt{\dfrac{m_c}{m_b}}$

$\dfrac{p_c}{p_b}=\sqrt{\dfrac{4.2\times 10^{-2}}{0.146}}$

So, $\dfrac{p_c}{p_b}=\dfrac{53}{100}$

So, the ratio of cardinal's magnitude of momentum to the magnitude of the baseball's momentum is 53 : 100

6 0

2 years ago

Who gains altitude more quickly, a pilot traveling 400 mph and rising at an angle of 30°, or a pilot traveling 300 mph and risin

Basile [38]

The rate of altitude increase is equivalent to the vertical component of the pilot's velocity. Let the first pilot's velocity be A and the second's be B.
Ay = Asin(∅)
Ay = 400sin(30)
Ay = 200 mph

By = Bsin(∅)
By = 300sin(40)
By = 192.8 mph

200 - 192.8 = 7.2 mph
The first pilot gains altitude faster by 7.2 mph than the second pilot.

4 0

2 years ago