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2 years ago

Which is NOT true of the Big Bang Theory?

Physics

1 answer:

grigory [225]2 years ago

3 0

Answer:

1. it explains what's happening im the universe now

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Say the tortoise and the hare are racing. If the tortoise moves straight

Rina8888 [55]

Explanation:

It's displacement would be negative

displacement is a vector quantity.

'Backwards', we can assume, would be negative.

and forwards, positive. So going backwards would mean a negative displacement.

5 0

3 years ago

A simple pendulum 2 m long swings through a maximum angle of 30 ? with the vertical.

Alexus [3.1K]

The complete question was calculate the period T assuming the smallest amplitude.
Using the equation;
T = 2 π√(L/g)
Where T is the period in seconds, L is the length of the rod or wire in meters and g is the acceleration due to gravity.
Hence; T = 2×3.14 × √(2/9.81)
= 6.28 × 0.4515
= 2.836 seconds

6 0

3 years ago

The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it

Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, $m=0.02 kg$

Kinetic energy imparted, $K=1200 J$

Length of rifle barrel, $d=1 m$

(a)

Let the speed of bullet when it leaves the barrel is $v$ .

Kinetic energy, $K=\frac{1}{2} mv^{2}$

$v=\sqrt{\frac{2K}{m} }$

$=\sqrt{\frac{2\times1200}{0.02} }$

$=346.4m/s$

Initial speed of bullet, $u=0$

The average speed in the barrel, $v_a_v_g=\frac{u+v}{2}$

$=\frac{0+346.4}{2} \\=173.2 m/s$

Time taken by bullet to cross the barrel, $t=\frac{d}{v}$

$=\frac{1}{173.2}\\ =0.00577 second$

Power, $P_a_v_g=\frac{W}{t}$

$=\frac{1200}{0.00577} \\=207.97kW$

(b)

In projectile motion,

Maximum height, $H_m=\frac{v^2\sin^2\theta}{2g} \\$

Range, $R=\frac{v^2\sin2\theta}{g}$

given that, $H_m=R$

then, $\frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter$

5 0

3 years ago

A simple pendulum has a period of 2.5 s. What is its period if its length is increased by a factor of four?

Svetach [21]

Answer:

Its period if its length is increased by a factor of four is 5 s.

Explanation:

The period of a simple pendulum is given by;

$T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g} \ be \ constant \\\\\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\$

Given;

initial period, T₁ = 2.5

initial length, = L₁

new length, L₂ = 4L₁

the new period, T₂ = ?

$\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\ T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\ T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s$

Therefore, its period if its length is increased by a factor of four is 5 s.

5 0

2 years ago

87) Determine the equivalent resistances for the following circuits.

stepladder [879]

Answer:

15 and 11 ohms

Explanation:

First one = For the parallel resistors 1 / (1/6 + 1/6 + 1/6) = 1/ (3/6 ) = 6/3 = 2 ohms then add the 3 and the 10 = 15 ohms

second one for the parallel portion equiv = (10+2)*24 / ( (10+2 + 24) = 8

then add the 3 in series = 11 ohms

8 0

2 years ago