Ask question

Ask question

All categories

11111nata11111 [884]

3 years ago

15

A coffee machine transfers 12 kj of energy in 15 seconds. It uses 230 v mains supply. Use this information to work out the curre

nt through the coffee machine

1 answer:

Otrada [13]3 years ago

6 0

Answer:

3.5 A

Explanation:

You might be interested in

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

Read 2 more answers

car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car

Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0

3 years ago

A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough h

ikadub [295]

Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

3 0

3 years ago

How was the gravitational constant G first determined

Oliga [24]

<span>In equation form, this is often expressed as follows: The constant of proportionality in this equation is G - the universal gravitation constant. The value of G was not experimentally determined until nearly a century later (1798) by Lord Henry Cavendish using a torsion balance.</span>

6 0

3 years ago

_____ have a nearly circular orbit.

bezimeni [28]

B. Asteroids have a nearly circular orbit.

4 0

3 years ago