A stone is thrown upward at an angle. what happens to the horizontal component of its velocity as it rises? as it falls?

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:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

5 0

3 years ago

The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90

s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0

3 years ago

Please help with this :(

Angelina_Jolie [31]

Answer:

I'm sorry I don't have a answer but I like your pfp

8 0

2 years ago

An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit

zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

$F=qE$

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

$q=3.0\cdot 10^{-6}C$

$E=-1.5\cdot 10^4 N/C$ (negative because inward)

So the strength of the electrostatic force is

$F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N$

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

6 0

3 years ago

Read 2 more answers

i just said something super cringey to someone and now i wish i had a time machine because i cant forget it :((( SOME HELP PLEAS

Llana [10]

Answer:

what did you say???

Explanation:

6 0

2 years ago

Read 2 more answers