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sveticcg [70]
3 years ago
14

A quarterback back pedals 3.3 meters southward and then runs 5.7 meters northward. For this motion, what is the distance moved?

What is the magnitude and direction of the displacement?
Physics
1 answer:
mixas84 [53]3 years ago
8 0

Answer:

The distance moved is 9 meters

The magnitude of the displacement is 2.4 meters

The direction of the displacement is northward

Explanation:

- <em><u>Distance</u></em> is the length of the actual path between the initial and the

  final position. Distance is a scalar quantity

- <u><em>Displacement</em></u> is the change in position, measuring from its starting

  position to the final position. Displacement is a vector quantity

The quarterback pedals 3.3 meters southward

That means it moves down 3.3 meters

Then runs 5.7 meters northward

That means it runs up 5.7 meters

The distance = 3.3 + 5.7 = 9 meters

<em></em>

<em>The distance moved is 9 meters</em>

It moves southward (down) for 3.3 meters and then moves northward

(up) for 5.7

It moves from zero to 3.3 down and then moves up to 5.7

The displacement = 5.7 - 3.3 = 2.4 meters northward

<em />

<em>The magnitude of the displacement is 2.4 meters</em>

<em>The direction of the displacement is northward</em>

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Calculate the drop voltage for a battery of internal resistance 2 ohm and emf of 24 volt
Ivenika [448]

Answer:

<em>The drop voltage is 0.3 V</em>

Explanation:

Electromotive Force EMF

When connecting a battery of internal resistance Ri and EMF ε to an external resistance Re, the current through the circuit is:

\displaystyle i=\frac{\varepsilon }{R_e+R_i}

The battery has an internal resistance of Ro=2 Ω, ε=24 V and is connected to an external resistance of Re=158 Ω. Thus, the current is:

\displaystyle i=\frac{24 }{158+2}

\displaystyle i=\frac{24 }{160}

i = 0.15 A

The drop voltage is the voltage of the internal resistance:

V_i = i.R_i

V_i = 0.15*2

\boxed{V_i = 0.3\ V}

The drop voltage is 0.3 V

3 0
3 years ago
While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta
Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to:  brainly.com/question/12550364

#SPJ4

8 0
2 years ago
4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer
Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

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What is meant by rusting​
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3 years ago
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