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3 years ago

Which is generally easier to accomplish? Prove a hypothesis Disprove a hypothesis

Physics

1 answer:

Iteru [2.4K]3 years ago

5 0

It's generally impossible to prove a hypothesis,
but one counterexample disproves it.

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Hallucinations, disorientation and the loss of a sense of time are symptoms of what disorder?-obsessive love disorder-alice in w

Fynjy0 [20]

The answer is Alzheimer's disease

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3 years ago

How much work is done when a 700 N refrigerator is lifted straight up a distance of 4 meters

KatRina [158]

Answer:

The work done is 2800 Joules

Explanation:

When a force causes a body to move, work is being done on the object

by the force

Work is the measure of energy transfer when a force moves an object

through a distance

→ Work = Force × Distance

→ The SI unit for work is the joule (J)

A refrigerator of 700 N is lifted straight up a distance of 4 meters

We need to find the work done

→ The weight of the refrigerator is 700 N

→ The force = 700 Newtons

→ The refrigerator is lifted straight up 4 m

→ The distance = 4 meters

Substitute these values in the rule below

→ Work = Force × Distance

→ Work = 700 × 4 = 2800 Joules

<em>The work done is 2800 Joules</em>

8 0

3 years ago

A bucket weighing 5 lbs is lifted at a constant rate from the bottom of a 100 ft well by a rope which weighs 5 lbs. The bucket h

8_murik_8 [283]

Answer:

$W_{bucket} = 24934.85\,lbf\cdot ft$

Explanation:

The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

$W_{bucket} = U_{g,B, bucket} - U_{g,A,bucket} +U_{g, B, water}-U_{g,A,water} +U_{g, B, rope} -U_{g,A,rope}$

$W_{bucket} = (5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-0\,ft) + (25\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft) - (30\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (0\,ft)+(5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-50\,ft)$

$W_{bucket} = 24934.85\,lbf\cdot ft$

6 0

4 years ago

Solving a series circuit, did I do this correctly?

nirvana33 [79]

The total resistance in the circuit is 16 Ohms.
The total current in the circuit is 0.5 Ampere.
The current at $R_1$ is 0.5 Ampere.
The current at $R_3$ is 0.5 Ampere.
The voltage drop at $R_1$ is 4 volts.
The voltage drop at $R_2$ is 2.5 volts.
The voltage drop at $R_3$ is 1.5 volts.
The total power consumed by the circuit is 4watts
The power consumed at $R_1$ is 2 watts
The power consumed at $R_2$ is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

$R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega$

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

The total resistance in the circuit = $R_{eq}$

$R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega$

The total current in the circuit = I

$V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A$ (Ohm's law)

For series combinations, the current in each resistor remains the same.

So, the current in $R_1, R_2 \&R_3$ :

$I_1= I_2= I_3=I=0.5 A\\$

The voltage drop across at $R_1$ = $V_1$

The current across $R_1$ = I = 0.5 A

$V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V$

The voltage drop across at $R_2$ = $V_2$

The current across $R_2$ = I = 0.5 A

$V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V$

The voltage drop across at $R_3$ = $V_3$

The current across $R_3$ = I = 0.5 A

$V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V$

The total power consumed by circuit:

$P= V\times I \\\\= 0.5 A\times 8 V = 4 watt$

Power consumed at $R_1$ :

$P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt$

Power consumed at $R_2$ :

$P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt$

Power consumed at $R_3$ :

$P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt$

Learn more about, current, voltage, resistance, and power of the circuit here:

brainly.com/question/11683246?referrer=searchResults

brainly.com/question/1430450?referrer=searchResults

5 0

2 years ago

A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t

dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

$P=m*v$

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

$(m_{1}*v_{1})=(m_{2}*v_{2})$

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

$(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]$

8 0

3 years ago