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3 years ago

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My question is An object may be acted on by several forces. What name is given to the single force that has the same effect as t

hese forces ?

2 answers:

12345 [234]3 years ago

8 0

Dat wuld be the Resultant Force

wariber [46]3 years ago

6 0

Newtons Law of motion

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Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the s

sveta [45]

Answer:

b

Explanation:

Given:

- The ball is fired at a upward initial speed v_yi = 2*v

- The ball in first experiment was fired at upward initial speed v_yi = v

- The ball in first experiment was as at position behind cart = x_1

Find:

How far behind the cart will the ball land, compared to the distance in the original experiment?

Solution:

- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:

v_yf = v_yi + a*t

Where, a = -9.81 m/s^2 acceleration due to gravity

v_y,f = 0 m/s max height for both cases:

For experiment 1 case:

0 = v - 9.81*t_1

t_1 = v / 9.81

For experiment 2 case:

0 = 2*v - 9.81*t_2

t_2 = 2*v / 9.81

The total time for the journey is twice that of t for both cases:

For experiment 1 case:

T_1 = 2*t_1

T_1 = 2*v / 9.81

For experiment 2 case:

T_2 = 2*t_2

T_2 = 4*v / 9.81

- Now use 2nd equation of motion in horizontal direction for both cases:

x = v_xi*T

For experiment 1 case:

x_1 = v_x1*T_1

x_1 = v_x1*2*v / 9.81

For experiment 2 case:

x_2 = v_x2*T_2

x_2 = v_x2*4*v / 9.81

- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.

- Hence; take ratio of two distances x_1 and x_2:

x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v

Simplify:

x_1 / x_2 = 2

- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.

3 0

3 years ago

Razona si, en las experiencias de Faraday, el efecto producido moviendo el circuito será el mismo que el producido moviendo el i

const2013 [10]

Answer:

moving the circuit or the magnet gives the same result

Explanation:

The faraday effect establishes that the temporal variation of imaginative flow produces an electric potential

fem = $\frac{d \phi }{dt}$ dfi / dt

the magnetic flux is

Ф = B. A = B A cos θ

suppose for simplicity that the angle is zero so cos 0 = 1

Φ = B A

By analyzing this expression, the change in magnetic flux can converge while keeping the magnetic field fixed and varying the electric field or keeping the electric field fixed and varying the magnetic field.

Consequently moving the circuit or the magnet gives the same result

5 0

3 years ago

I was playing goalie in a soccer league and made a save. The ball was traveling at 14 m/s and has a mass of 0.45 kg. How muck ki

Zigmanuir [339]

Answer:

Kinetic energy = 44.1 J

Explanation:

Given:

Velocity of ball = 14 m/s

Mass of ball = 0.45 kg

Find:

Kinetic energy

Computation:

An item's kinetic energy is force it has as a result of its motion. It is the amount of work necessary to move a body of a known volume from rest to a specified velocity.

Ke = (1/2)(m)(v²)

Ke = (1/2)(0.45)(14²)

Ke = (1/2)(0.45)(196)

Kinetic energy = 44.1 J

6 0

3 years ago

Two cars A and B are 100m apart moving towards each other with

maxonik [38]

Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0

3 years ago

g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b

Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c) Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

Em₀ = K = ½ m v²

final point. Higher

Em_f = U = mg h

Em₀ = Em_f

Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

Em₁ = K = ½ m v²

final point. Higher

Em_f = U = mg h

Em₁ = Em_f

Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0

3 years ago