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2 years ago

I will mark you brainlist!

Physics

1 answer:

kirill115 [55]2 years ago

7 0

Tornado- Trees knocked down, debris everywhere, ground and dirt scattered.

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A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels

Kryger [21]

Answer:

T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

$T = \dfrac{v^2.m}{l}$

$v = \dfrac{l}{s}$

$v = \dfrac{7.2}{0.74}$

v = 9.73 m/s

now, calculation of tension

$T = \dfrac{9.73^2\times 0.46}{7.2}$

T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0

2 years ago

Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re

Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0

2 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

Some homes that use baseboard heating use copper tubing. hot water runs through and heats the copper tubing, which in turn heats

AlekseyPX

When you heat a certain substance with a difference of temperature $\Delta T$ the heat (energy) you must give to it is
$E(=Q) =mc\Delta T$
where $c$ is the specific heat of that substance (given in J/(g*Celsius))
In this case
$E=645*0.3850*(28.22-13.20) =3729.8 (Joule)$

Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin) because on the temperature scale a difference of 1 degree Celsius = 1 degree Kelvin

7 0

2 years ago

1. The term that describes where the supply curve intersects the demand curve is known as

nadezda [96]

Equilibrium is the answer

6 0

2 years ago