Please help!!!!!!!!!

A block of metal which is 20 cm on a side has a mass of 20 kg. What is it’s density? Please answer in MkS (SI) units.

Ratling [72]

The density of an object is given by:

D = M/V

D = density, M = mass, V = volume

The volume of the cube V is given by:

V = s³

where s = side length

Make a substitution:

D = M/s³

Given values:

M = 20kg, s = 20×10⁻²m

Plug in and solve for D:

D = 20/(20×10⁻²)³

D = 2500kg/m³

6 0

4 years ago

This equation represents Newton's Second Law of Motion, F=ma. It shows the relationship of force, mass, and acceleration. How wi

inn [45]

A. The acceleration will be doubled

3 0

3 years ago

The maximum potential energy of a spring system (mass 15 kg, spring constant 850 N/m) is 6.5 J. a) What is the amplitude of the

Bumek [7]

Answer:

a) 0.124 m

b) 0.93 ms⁻¹

c) 0.5 k A² cos ² ( ωt )

Explanation:

1) Potential energy = U = 0.5 k A² , where A is the amplitude and k = 850 N/m is the spring constant.

0.5 ( 850) (A² ) = 6.5

⇒ A = 0.124 m = Amplitude.

b) From energy conservation, 0.5 m v² = 6.5

⇒ speed = v = 0.93 ms⁻¹

c) If x = A cos ωt ,

Potential energy = 0.5 k A² = 0.5 k A² cos ² ( ωt )

8 0

3 years ago

you throw a ball with a mass of 2.1 kg. The balls leave your hand at 30 m/s the ball has_________ enrgy. Calculate it

Murrr4er [49]

Answer:

945 joules

Explanation:

Kinetic energy = 1/2(MV²)

Given parameters

Mass = 2.1kg

Velocity = 30m/s

K.E = 1/2(2.1 x 30²)

K.E = 1/2( 2.1 x 900)

K.E = 2.1 x 450

K.E = 945 joules

7 0

3 years ago

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on

Allushta [10]

To solve the exercise it is necessary to apply the equations necessary to apply Newton's second law and the concept related to frictional force.

An angle of 30 degrees is formed on the vertical at an applied force of 2.3N

In this way the frictional force, opposite to the movement will be given by

$f_k = \mu_k N$

where,

$\mu_k =$ Kinetic friction constant

N = Normal Force (Mass*gravity)

The friction force is completely vertical and opposes the rising force of 2.3 N. The Normal force acts perpendicular to the surface (vertical) therefore corresponds to the horizontal component of the applied force.

The ascending force would be given by

$F_v = 2.3N*Cos30 = 1.99N$

As the block is moving upward, the friction force acts downward, also its weight acts downward. We can have

$2.3N+f_k = 1.99N$

$f_k = 0.31N$

Considering the horizontal force the normal force on the block must be balanced by the horizontal component of pishing foce

$N = 2.3sin30$

$N = 1.15N$

Then the frictional force

$f_k = \mu_k N$

$0.31N = \mu_k 1.15$

$\mu_k = \frac{0.31}{1.15}$

$\mu_k =0.26$

Therefore the coefficient of kinetic friction is 0.26

3 0

3 years ago