How many times does the light beam shown in FIGURE 26-59 reflect from (a) the top and (b) the bottom mirror?

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

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#LearnwithBrainly

5 0

3 years ago

You drive 6.00 km at 50.0 km/h and then another 6.00kmat 900 km/h Your average speed over

Cerrena [4.2K]

Explanation:

average speed = total distance travelled / total time travelled

time to travel the first 6km: 6 / 50 = 3/25 (h)

time to travel the next 6km: 6 / 90 = 1/15 (h)

[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]

The total time: 3/25 + 1/15 = 14/75 (h)

Average speed over the 12 km drive will be:

$\frac{12}{ \frac{14}{75} } = \frac{450}{7} = 64.3 \: km{h}^{ - 1}$

8 0

2 years ago

A circular wire loop of radius 15.0 cm carries a current of 2.60

Korolek [52]

Part (a): Magnetic dipole moment

Magnetic dipole moment = IA, I = Current, A = Area of the loop
Then,
Magnetic dipole moment = 2.6*π*0.15^2 = 0.184 Am^2

Part (b): Torque acting on the loop
T = IAB SinФ, where B = Magnetic field, Ф = Angle
Then,
T = Magnetic dipole moment*B*SinФ = 0.184*12*Sin 41 = 1.447 Nm

5 0

3 years ago

RANGKAIAN RLC SERI PADA TEGANGAN BOLAK BALIK 220V R=80 OHM XL= 90 OHM XC = 30 OHM MAKA TEGANGAN PADDA INDUKTOR ADALAH???

Whitepunk [10]

Okay, 90% of this is nonsense besides the numbers maybe.

8 0

3 years ago

A 1060-kg car moving west at 16 m/s collides with and locks onto a 1830-kg stationary car. determine the velocity of the cars ju

iren [92.7K]

M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s

5 0

3 years ago