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kherson [118]
3 years ago
12

An object falls for 1.7 seconds. Assuming uniform acceleration, how far has it moved? ​

Physics
1 answer:
anastassius [24]3 years ago
6 0

It has moved

(1.445) • (the acceleration) meters

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If a CD player uses a 12-V battery and draws 2.0 amps of current, how much power does it use?
IgorC [24]

Answer:

The power is 24 watt .

Explanation:

Given that,

Voltage V= 12-V

Current I = 2.0 A

Using ohm's law

The current is directly proportional to the voltage.

In form of power,

The power is the product of the Current and voltage.

Formula of the power is defined as,

P = I\times V

P = 2.0\times12

P = 24\ watt

Hence, The power is 24 watt .

4 0
2 years ago
I need help ASAP plzzzz
Fiesta28 [93]

Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

a= -9.81 m/s^2 t=???

Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

-15mx2/a= t^2

Square root (-30/-9.81m/s^2)

t=1.75 s

b) now v2!!

Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

v2 = -17.2 m/s or you can say 17.2 m/s down!!!

7 0
2 years ago
When you enter a toll road, your ticket is stamped 1:00 p.m. When you leave, after traveling 55 miles, your ticket is stamped 2:
Viefleur [7K]
27.5 because of you divide the 55miles with the time you get your velocity which is the speed.
4 0
2 years ago
Read 2 more answers
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
3 years ago
How does the height from which you drop the ball relate to the height that the ball bounces back up?
Stels [109]
The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.
7 0
3 years ago
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