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GuDViN [60]
2 years ago
14

two cars start at the same point and drive in a straight line for 5km. At the end of the drive their distances are the same but

their displacements are different. Explain how this could happen. Illustrate with a diagram.
Physics
1 answer:
Anna11 [10]2 years ago
7 0

A 'displacement' always consists of a magnitude and a direction.  The two cars you just described have displacements with the same magnitude ... 5 km.  But if they didn't both drive in the same direction, then their displacements are different.

Remember:

-- 10 m/s² up and 10 m/s² down are different accelerations

-- 30 mph East and 30 mph West are the same speed but different velocity.

-- 5 km North and 5 km South are the same distance but different displacement.

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A 6.00 A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5×1028 free elect
navik [9.2K]

Answer:

Explanation:

Current, I = 6 A

diameter of wire, d = 2.05 mm

number of electrons per unit volume, n = 8.5 x 10^28

If the diameter is doubled,

The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is  one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.  

8 0
3 years ago
Find the velociity of a car which travels 35 m to the right over a period of 40 seconds
GalinKa [24]

Answer:

the velocity of the car is 0.875 m/s

Explanation:

v = d \div t \\ v = 35m \div 40sec = 0.875 \: m \: per \: sec

therefore the V of car is 0.875 m

8 0
3 years ago
A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem
Shtirlitz [24]

Answer:

U = (ε0AV^2) / 2d

Explanation:

Where C= capacitance of the capacitor

ε0= permittivity of free space

A= cross sectional area of plates

d= distance between the plates

V= potential difference

First, the capacitance of a capacitor is obtained by:

C = ε0A/d.

Starting at the formula , U= (CV^2)/2. Formula for energy stored in a capacitor

Substitute in for C:

U = (ε0A/d) * V^2 / 2

Hence:

U = (ε0AV^2) / 2d

3 0
2 years ago
A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the
Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0
3 years ago
Which has the most momentum?
boyakko [2]

Answer:

Both objects have the same magnitude of momentum.

Explanation:

If an object of mass m is moving at a velocity of v, the momentum of that object would be m\, v.

The 100\; {\rm g} (0.1\; {\rm kg}) object is moving at a speed of 1\; {\rm m\cdot s^{-1}}. The magnitude of the momentum of this object would be 0.1\; {\rm kg} \times 1\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}.

Similarly, the momentum of the 1\; {\rm g} (10^{-3}\; {\rm kg}) object moving at a speed of 100\; {\rm m\cdot s^{-1}} would be 10^{-3}\; {\rm kg} \times 100\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}.

Hence, the magnitude of momentum is the same for the two objects.

7 0
1 year ago
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