I found an answer from www.bbc.com

Weight, mass and gravitational **field** strength - Newton's Laws ...

Example. An apple has a mass of 100 g. Calculate its weight on Earth (g = **10** N/
kg). 100 g = 100 ÷ 1000 = 0.1 kg.

For more information, see Weight, mass and gravitational **field** strength - Newton's Laws ...

I found an answer from www.quora.com

A water tank is **3 m** long, **2 m** wide and **1 m** high. How many litres ...

Given; **length**=3m, breadth=2m, **height**=1m. Since **volume** of a regular cuboid=
**length**×breadth×**height** So, **volume**=3m×2m×1m **Volume**=6m^**3** Now, since ...

For more information, see A water tank is **3 m** long, **2 m** wide and **1 m** high. How many litres ...

I found an answer from www.examfear.com

CBSE **NCERT** Solution for Class 11 - **Physics** - **Units** and ...

CBSE **Ncert** Solution for Class 11 - **Physics** - **Units** and **Measurements**. ... (a) The
**volume** of a **cube** of side 1 cm is equal to .....m^{3} ... The **total surface area** of a
**cylinder** of radius r and height h is: ... **Given Time**, t = 1sec. ... Suppose we employ
a system of **units** in which the **unit** of **mass** equals α kg, the **unit** of **length** equals β
m ...

For more information, see CBSE **NCERT** Solution for Class 11 - **Physics** - **Units** and ...

I found an answer from www.scientificamerican.com

The Bicycle Problem That Nearly Broke **Mathematics** - Scientific ...

Jul 20, 2016 **...** Jim Papadopoulos has spent a lifetime pondering the **maths** of bikes in motion. ...
He could never ride a bike without pondering the **mathematical** mysteries that it
contained. ... Why must **one** initially steer right in order to lean and turn left? ...
anywhere from 5 **centimetres** to **10 centimetres** behind the steering ...

For more information, see The Bicycle Problem That Nearly Broke **Mathematics** - Scientific ...

a)

Formula volume of the cube = x^3 cubic units

Given that, side of the cube x = 1cm

Volume of the cube = 1 cm * 1 cm * 1 cm = 1 cm^3

Hence, cube volume V = 1 cm^3 = 1* 10^{-6} m^3

b)

Radius of the cylinder r = 2.0 cm

Height of the cylinder h = 10.0 cm

Surface area of the cylinder S.A = Curved Surface + Area of Circular bases

S.A = 2\pi r(h + r) square units

Plug in the given values in the formula

S.A = 2 * 3.14 * 2.0(10 + 2.0)

S.A = 150.72 cm^2

S.A = 15072 mm^2 \because 1 cm = 10 mm

c)

Speed v = 18 km/h = 18* \frac{5}{18} m/s = 5 m/s \because 1 m/s = \frac{1000}{60* 60} = \frac{5}{18} m/s

Time t = 1 seconds

\text{ speed } = \frac{\text{ Distance }}{\text{ time }}

\text{ Distance } = speed * time = 5 m/s * 1 s = 5 m

Thus, distance covered d = 5 m

d)

Lead relative density RD = 11. 3

Density of lead \rho = ?

Density of the water(reference) \rho_w = 1 g/cm^3 = 1000 kg/m^3

Formula: \text{relative density substance} = \frac{\text{ density of substance }}{\text{density of reference }}

\text{ density of substance } = \text{ relative density substance } * \text{ density of reference }

\text{ density of substance } = 11.3 * 1000 kg/m^3

\text{ density of substance } = 11300 kg/m^3