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3 years ago

ruto is 12 years old.in three years time he will be ⅓ of his father present age.how old was his father 12 years ago

Physics

1 answer:

max2010maxim [7]3 years ago

4 0

Answer: 33

Explanation:

12+3=15, so in 3 years time Ruto will be 15.

Because 15 is said to be 1/3 times his fathers present age (x), you can plug that into an equation.

15=1/3(x)

Then solve the equation.

(3)15=[1/3(x)]•3

45=x

His father is currently 45, so if you subtract 12 from that, you get 33.

Ruto’s father was 33 years old 12 years ago.

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Which scale has 100 divisions from when the temperature when water freezes to the temperature when water boils

Westkost [7]

The Celsius scale ( $^{\circ} C$ ).

In the Celsius temperature scale, the temperature at which water freezes is set conventionally at $0^{\circ}C$ , while the temperature at which the water boils is set at $100^{\circ}C$ . The Celsius degree is then defined as the unit corresponding to 1/100 of this time interval, between the temperature of freezing and boiling of the water.

8 0

3 years ago

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

daser333 [38]

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

$T = FR = 18 * 2.4 = 43.2 Nm$

According to Newton 2nd law, the angular acceleration would be

$\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2$

It starts from rest, then after 15s it would achieve a speed of

$\omega = \alpha t = 0.021 * 15 = 0.31 rad/s$

(b) The distance angle swept by it is:

$\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad$

Hence the work by the child

$W = T\theta = 43.2 *2.314 \approx 100 J$

c) Average power to work per time unit

$P = \frac{W}{t} = \frac{100}{15} = 6.67 W$

7 0

3 years ago

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a

Ilya [14]

Answer with Explanation:

We are given that

Area of loop= $(20\times 20) cm^2=400\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Resistance, R= $0.1\Omega$

B= $4t-2t^2$

We know that magnetic flux

$\phi=BA$

Emf , $E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid$