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1 year ago

A shopping mall is putting up lights for the holidays. When they plug the holiday lights in they do not light up.

Physics

1 answer:

mash [69]1 year ago

3 0

Answer:

Explanation:

Remark

Anybody who has dealt with Christmas Tree lights knows what can be done. You must take a bulb you know to be good and try each bulb in turn until the whole string lights up. If two bulbs are dead, give up and go buy another string. Two bulbs create a lot of combinations.

The circuit is a series circuit.

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In Milgram's experiment:

SpyIntel [72]

Answer:

B. The "Learner" was working with Milgram.

Explanation:

just took the test

give brainliest, please. :)

3 0

2 years ago

A meter stick A hurtles through space at a speed v = 0.25c relative to you, with its length aligned with the direction of motion

yaroslaw [1]

Answer:

$L_0\approx1.0328\ m$

Explanation:

Given:

relativistic length of stick A, $L=1\ m$

relativistic velocity of stick A with respect to observer, $v=0.25c=7.5\times 10^{7}\ m.s^{-1}$

<em>Since the object is moving with a velocity comparable to the velocity of light with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>

<u> Mathematical expression of the theory of relativity for length contraction:</u>

$L=\frac{L_0}{\gamma}$

where:

L = relativistic length

$L_0=$ original length at rest

$\gamma =$ Lorentz factor $=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$

$\Rightarrow 1=\frac{L_0}{\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }}$

$L_0=\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }$

$L_0\approx1.0328\ m$

4 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

2 years ago

You are working in the finance department of Innotech Ltd (INT). The Company has spent $5 million

lozanna [386]

Answer: The Company has spent $5 million in research and development over the past 12 months developing cutting-edge battery technology which will be incorporated ...

Explanation: uhmmmmmm i dont know this one but it is pretty ez

5 0

3 years ago

An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator i

avanturin [10]

Answer: Tension = 53.6N

Explanation:

Given that

Height h = 1 m

Time t = 1.7 s.

Mass m = 5.1 kg

From the equation of the motion we can get the acceleration of the elevator:

h = X0+ V0t + at2/2;

Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus

a = 2h/t2 = 2 × 1/1.7^2

a = 0.69 m/s2.

Then we can find the tension in the cord by using the formula

T = mg + ma

= 5.1 (9.8 + 0.69)

= 5.1 × 10.5

= 53.6N

7 0

3 years ago