This direct relationship graph shows that the position of the car as time elapses.

What is your least favorite candy? Explain

aev [14]

Answer:

My least favorite is whoppers.....Trust me i love chocolate, but not when it taste like chalk......

7 0

2 years ago

Read 2 more answers

A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor

Mumz [18]

Answer:

The maximum induced emf in the rotating coil = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

$emf_max$ = NBAω ; if (A = l × w) , we have:

$emf_max$ = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

$emf_max$ = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e $emf_max$ = $N\frac{d∅}{dt}$

since $emf_max$ = 29.66 and N = 44; we have:

29.66 = $44\frac{d∅}{dt}$

$\frac{d∅}{dt}$ = $\frac{29.66}{44}$

= 0.674 Wb/s

5 0

3 years ago

A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound l

DENIUS [597]

Answer:

$\mathbf{\beta = 123.75 \ dB}$

Explanation:

From the question, using the expression:

$125 \ dB = 10 \ log (\dfrac{I}{I_o})$

where;

$I_o = 10^{-12} \ W/m^2$

$I = 10^{12.5} \times 10^{-12} \ W/m^2$

$I = 3.162 \ W/m^2$

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = $\dfrac{3.162\times 3}{4}$

= 2.372 W/m²

Thus;

$\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2$

$\mathbf{\beta = 123.75 \ dB}$

7 0

2 years ago

A .5 kg toy train car moving forward at 3 m/s collides with and sticks to a .8 kg toy car that is traveling at 2 m/s what is the

Viktor [21]

Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
$m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1}+ m_{2} * v'_{2}$

In case of perfectly inelastic collision v'1 and v'2 are same.

We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x

0.5*3 + 0.8*2 = 0.5*x + 0.8*x
1.5 + 1.6 = 1.3x
3.1 = 1.3x
x = 2.4 m/s

4 0

2 years ago

A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90

Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115= 3.175

now,

$a= \sqrt{a_t^2+a_r^2}$

$a= \sqrt{1.47^2+3.175^2}$

a= 3.49 m/s^2

3 0

2 years ago