A body accelerates uniformly from rest at 2ms² for 5 seconds . calculate its average velocity in this time.

Determinar el flujo de calor a través del piso de losa cuyas medidas 3 X 5 cm y temperaturas superficiales son -20 ºC y 40 ºC, l

sweet-ann [11.9K]

Answer:

0,0560 cal / gºC.

3 0

3 years ago

The position (in radians) of a car traveling around a curve is described by Θ (t) = t 3 - 2t 2 - 4t + 10 where t (in seconds). W

FromTheMoon [43]

Answer: $\alpha =30-4=26 rad/s^2$

Explanation:

Given

Position of a car is given by

$\theta =t^3-2t^2-4t+10$

and angular speed is $\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

angular acceleration is

$\frac{\mathrm{d^2} \theta }{\mathrm{d} t^2}=\alpha$

$\alpha =6t-4$

at $t=5 s$

$\alpha =30-4=26 rad/s^2$

4 0

4 years ago

What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two

lesya [120]

Answer:

Power = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

$\beta (dB)= 10 \ log_{10 } (\dfrac{I}{I_o})$

where;

the threshold of hearing $I_o = 10^{-12} (W/m^2)$

$\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})$

$10^{10.2} = \dfrac{I}{10^{-12}}$

$I = 10^{10.2} \times 10^{-12}$

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power = 124.50 W

7 0

3 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

7 0

3 years ago

A transformer is designed to provide 6 V from a 150 V supply. If the primary coil has 1000 turns, how many turns does the second

zavuch27 [327]

The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.

Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.

Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.

7 0

3 years ago

A body accelerates uniformly from rest at 2ms² for 5 seconds . calculate its average velocity in this time.​

A body accelerates uniformly from rest at 2ms² for 5 seconds . calculate its average velocity in this time.