Which of the following is equivalent to 10 kilograms?

28. Ken and Musa shared a cake such that Ken got twice the size

ratelena [41]

Answer:

$Musa = \frac{1}{3}$

$Ken = \frac{2}{3}$

Explanation:

Given

$Ken = 2 * Musa$ --- Ken's share

Required

The fraction each got

Since they both shared a cake, we have:

$Ken + Musa = 1$

Substitute: $Ken = 2 * Musa$

$2 * Musa+ Musa = 1$

Factorize

$Musa(2+ 1)= 1$

$Musa(3)= 1$

Divide both sides by 3

$Musa = \frac{1}{3}$

Recall that: $Ken = 2 * Musa$

$Ken = 2 * \frac{1}{3}$

$Ken = \frac{2}{3}$

3 0

3 years ago

Which is an SI base unit that makes up part of the unit of force?

Snezhnost [94]

Answer:

Which is an SI base unit that makes up part of the unit of force?

1.candela

5 0

3 years ago

A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use the an

lana66690 [7]

Answer:

d= 64.7 km

$\theta = 40.9^{o}$

displacement vector $=r_xi + r_yj = 48.9i + 42.4j$

Explanation:

total distance = 40 + 30 = 70 km

during 1st flight

$r_1 x = 40*cos60$

$r_1 x = 20 km$

$r_1 y = 40*sin60$

$r_1 y = 34.64 km$

during 2nd flight

$r_2 x = 30*cos15$

$r_2 x = 28.9 km$

$r_2 y = 30*sin15$

$r_2 y = 7.76 km$

the two component of r are:

$r_x = r_1x + r_2x = 20 + 28.9 = 48.9 km$

$r_y = r_1y + r_2y = 34.64 + 7.76 = 42.4 km$

Geographical Direction $\theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}$

Displacement d $= \sqrt{r_x^2 + r_y^2}$

$d = \sqrt{48.9^2+42.4^2} = 64.7 km$

d= 64.7 km

displacement vector $=r_xi + r_yj = 48.9i + 42.4j$

8 0

3 years ago

Col. John Stapp led the U.S. Air Force Aero Medical Laboratory's research into the effects of higher accelerations. On Stapp's f

Lerok [7]

Answer:

A. $a=203.14\ \frac{m}{s^2}$

B.s=397.6 m

Explanation:

Given that

speed u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

0 = 284.4 -a x 1.4 (here we take negative sign because this is the case of de acceleration)

$a=203.14\ \frac{m}{s^2}$

So the acceleration while stopping will be $a=203.14\ \frac{m}{s^2}$ .

Lets take distance travel before come top rest is s

We know that

$v^2=u^2-2as$

$0=284.4^2-2\times 203.14\times s$

s=397.6 m

So the distance travel while stopping is 397.6 m.

8 0

4 years ago

Read 2 more answers

A loop of wire in the shape of a rectangle rotates with a frequency of 284 rotation per minute in an applied magnetic field of m

yaroslaw [1]

Answer:

a) Magnitude of maximum emf induced = 0.0714 V = 71.4 mv

b) Maximum current through the bulb = 0.00793 A = 7.93 mA

Explanation:

a) The induced emf from Faraday's law of electromagnetic induction is related to angular velocity through

E = NABw sin wt

The maximum emf occurs when (sin wt) = 1

Maximum Emf = NABw

N = 1

A = 4 cm² = 0.0004 m²

B = 6 T

w = (284/60) × 2π = 29.75 rad/s

E(max) = 1×0.0004×6×29.75 = 0.0714 V = 71.4 mV

Note that: since we're after only the magnitude of the induced emf, the minus sign that indicates that the induced emf is 8n the direction opposite to the change in magnetic flux, is ignored for this question.

b) Maximum current through the bulb

E(max) = I(max) × R

R = 9 ohms

E(max) = 0.0714 V

I(max) = ?

0.0714 = I(max) × 9

I(max) = (0.0714/9) = 0.00793 A = 7.93 mA

Hope this Helps!!

8 0

3 years ago