What is a circuit that only has one loop??

A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially

zysi [14]

Answer:

x(t) = -8sin2t

Explanation:

See the attachment for solution

From my solving, we can deduce that w² = 4, and thus, w = 2

Therefore, the general solution is

x(t) = c1 cos2t + c2 sin2t

Considering the final variable, we can conclude that

x(0) = 0

x'(0) = -8 m/s

The final solution, thus

x(t) = -8sin2t

4 0

2 years ago

Kristen is running a half marathon. By her tenth minute of running, her breathing rate has increased from her resting rate.

pashok25 [27]

They need more oxygen to function, so gas exchange needs to increase

7 0

3 years ago

Read 2 more answers

A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o

Tema [17]

Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

W = P Δ V

W = 1.013 x 10⁵ x (19.7 - 13.7)

W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

now,

change in internal energy

Δ U = Q - W

Q = 3300 x 10³ cal

Q = 3300 x 10³ x 4.186 J

Q = 1.38 x 10⁷ J

now,

Δ U = 1.38 x 10⁷ - 6.029 x 10⁵

Δ U = 1.32 x 10⁷ J

6 0

3 years ago

An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b

trapecia [35]

Answer: Option B: 1.3×10⁵ W

Explanation:

$Power = \frac{Work \hspace{1mm} done}{Time}$

$P=\frac{W}{t}$

Work Done, $W= F.s$

Where s is displacement in the direction of force and F is force.

$\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v$

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

6 0

3 years ago

Calculate the pressure exerted on the heel of a boy’s foot if the boy weighs 80 N and he lands on one heel,which has an area of

cricket20 [7]

Pressure at a given surface is given as ratio of normal force and area

so here force due to heel of the shoes is given as 80 N

and the area of the heel is given as 16 cm^2

so we can say

$P = \frac{F}{A}$

here we have

F = 80 N

$A = 16 cm^2 = 16 * 10^{-4} m^2$

$P = \frac{80}{16 * 10^{-4}}$

$P = 5 * 10^4 N/m^2$

so pressure at the surface due to its heel will be 5 * 10^4 N/m^2

3 0

3 years ago