A clock radio is rated as 30 w of power output. if the radio also draws 30 w at 120 v, which will the current draw be?

A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4

Vedmedyk [2.9K]

Hello!

Vx = V0x + Ax*t
Vx = 18.1 + 2.4t

Let’s take time as 7.50 seconds:
Vx = 18.1 + 2.4*7.50
Vx = 18.1 + 18 = 36.1 m/s

Then, the final velocity of the car is 36.1 m/s.

3 0

3 years ago

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from

krek1111 [17]

Answer:

$E=35921.96N/C$

Explanation:

From the question we are told that:

Radius $r=0.321mm$

Charge Density $\mu=0.100$

Distance $d= 5.00 cm$

Generally the equation for electric field is mathematically given by

$E=\frac{mu}{2\pi E_0r}$

$E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}$

$E=35921.96N/C$

4 0

3 years ago

Additional Problem: A simple pendulum, consisting of a string (of negligible mass) of length L with a small mass m at the end, i

kykrilka [37]

Answer:

a) v = √ 2gL abd b) θ = 45º

Explanation:

a) for this part we use the law of conservation of energy,

Highest starting point

Em₀ = U = mg h

Final point. Lower

Em₂ = ½ m v²

Em₀ = Em₂

m g h = ½ m v²

v = √2g h

v = √ 2gL

b) the definition of power is the relationship between work and time, but work is the product of force by displacement

P = W / t = F. d / t = F. v

If we use Newton's second law, with one axis of the tangential reference system to the trajectory and the other perpendicular, in the direction of the rope, the only force we have to break down is the weight

sin θ = Wt / W

Wt = W sin θ

This force is parallel to the movement and also to the speed, whereby the scalar product is reduced to the ordinary product

P = F v

The equation that describes the pendulum's motion is

θ = θ₀ cos (wt)

Let's replace

P = (W sin θ) θ₀ cos (wt)

P = W θ₀ sint θ cos (wt)

We use the equation of rotational kinematics

θ = wt

P = Wθ₀ sin θ cos θ

Let's use

sin 2θ = 2 sin θ cos θ

P = Wθ₀/2 sin 2θ

This expression is maximum when the sine has a value of one (sin 2θ = 1), which occurs for 90º,

2θ = 90

θ = 45º

5 0

3 years ago

A rock is thrown at an angle of 30 degrees above the horizontal with initial velocity 15m/s what is the displacement when the ro

Delvig [45]

The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by

$x=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\cos30^\circ\,t$

so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:

$y=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\sin30^\circ\,t-\dfrac g2t^2=0$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. Then we find that $t\approx1.5\,\mathrm s$ , at which point we find $x\approx20\,\mathrm m$ .

6 0

3 years ago

An arrow is shot from a height of 1.55 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° abo

Mumz [18]

Answer:

Explanation:

vertical component of the velocity of arrow

= 26 sin 60 = 22.516 m

height reached by it after 3.99 s

h = ut - 1/2 g t²

= 22.516 x 3.99 - .5 x 9.8 x 3.99²

= 89.83 - 78

11.83 m

Total height of cliff = 1.55 + 11.83

= 13.38 m

c ) maximum height covered s

v² = u² - 2gs

0 = u² - 2gs

s = u² / 2g

= 22.516² / 2 x 9.8

= 25.86

maximum height reached

= 25.86 + 1.55

= 27.41 m

d )

vertical speed after 3.99 s

v = u - gt

= 22.516 - 9.8 x 3.99

= -16.586

Horizontal component will remain unchanged

Horizontal component = 26 cos 60

= 13 m /s

Resultant of two velocities

= √ 13²+ 16.568²

= 21 m /s

3 0

3 years ago