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3 years ago

Please help im stuck and can't find the answer sheet ANYWHERE :( PLEASEEEEEEEEEEE

Physics

1 answer:

Vladimir [108]3 years ago

6 0

Answer:

All steps are 20 * 100 (break the rest into appropriate pieces)

You can multiply as follows

(2000) * ((3 * 60) + (2 * 60) + 60)

V = 2000 * 6 * 60) = 720,000 cm^3 = .72 m^3

.72 m^3 * 2400 kg / m^3 = 1728 kg

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A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released

anastassius [24]

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

substituting the values of the variables into the equation, we have

k = mg/x

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

5 0

3 years ago

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the ma

Law Incorporation [45]

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration, $a_t$ = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of tangential acceleration and radial acceleration

$a = \sqrt{a_c^2 + a_t^2}\\\\$

where;

$a_c$ is the radial acceleration

$a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2$

The radial acceleration relates to speed of particle in the following equations;

$a_c = \frac{v^2}{r}$

where;

v is the speed of the particle

$v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s$

Therefore, the speed of the particle is 2.86 m/s

6 0

4 years ago

Consider two identical insulated metal spheres, A and B. Sphere A initially has a charge of -6.0 units and sphere B initially ha

Oksi-84 [34.3K]

Answer:

<em>-2 units of charge</em>

Explanation:

charge on A = Qa = -6 units

charge on B = Qb = 2 units

if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.

charge on each sphere will be = $\frac{Qa + Qb}{2}$

charge on each sphere = $\frac{-6 + 2}{2}$ = $\frac{-4}{2}$ = <em>-2 units of charge</em>

8 0

3 years ago

A refrigerator has a coefficient of performance equal to 4.00. The refrigerator takes in 110 J of energy from a cold reservoir i

Helga [31]

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

$Q_c = 110 \ J$

Coefficient of performance refrigerator,

$Cop(refrig)=4$

(a)

As we know,

⇒ $Cop(refrig) = \frac{Q_c}{Work}$

or,

⇒ $Work=\frac{Q_c}{Cop(refrig)}$

$=\frac{110}{4}$

$=27.5 \ Joules$

(b)

⇒ $Heat \ expelled = Heat \ removed +Work \ done$

or,

⇒ $Q_h = Q_c+Work$

$=114+27.5$

$=141.5 \ Joules$

4 0

3 years ago

What is the force between two proton’s that are 0.005m apart?

Jet001 [13]

Answer:

Force, $F=9.21\times 10^{-24}\ N$

Explanation:

Distance between proton is 0.005 m

It is required to find the force between two protons. Protons have of positive charge of $1.6\times 10^{-19}\ C$ . Two protons will have a force of repulsion between them. The force is given by :

$F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.005)^2}\\\\F=9.21\times 10^{-24}\ N$

So, the force between two protons is $9.21\times 10^{-24}\ N$ .

7 0

4 years ago

Please help im stuck and can't find the answer sheet ANYWHERE :(​ PLEASEEEEEEEEEEE

Please help im stuck and can't find the answer sheet ANYWHERE :( PLEASEEEEEEEEEEE