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satela [25.4K]
4 years ago
5

During a 400-m race, a runner crosses the 100 m mark with a velocity of 12 m/s. What would be her final position if she maintain

ed her velocity for another 10 seconds (in m)?

Physics
2 answers:
Cerrena [4.2K]4 years ago
7 0

Answer:

Explanation:

Given that

Total race distance is 400m

Her initial velocity was 0m/s²

At the 100m mark, after she has travelled 100m, her final velocity was v=12m/s²

Using equation of motion

Let determine her constant acceleration

v²=u²+2as

12²=0²+2×a×100

144=0+200a

144=200a

a=144/200

a=0.72m/s²

Then we want to know her position after another 10second

So total time is 10+12=22seconds

Then, using equation of motion

Let determine his postion

S=ut+½at²

S=0•t+½×0.72×22²

S=0+174.24

S=174.24 m

Her position will be 174.24m

Ratling [72]4 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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Allushta [10]
The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
p_i = p_1-p_2 = mv-mv =0
So, the final momentum must be zero as well:
p_f = 0
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
p_f = mv_1 + mv_2 = 0
From which
v_1 = -v_2

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is \frac{1}{2} mv^2. Therefore, the total kinetic energy before the collision is
K_i = \frac{1}{2} mv^2 +  \frac{1}{2} mv^2 = mv^2
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
K_f = K_i = mv^2 (1)
But the final kinetic energy, Kf, is also
K_f =  \frac{1}{2} mv_1^2 +  \frac{1}{2}mv_2^2
Substituting v_1 = -v_2 as we found in the conservation of momentum, this becomes
K_f = mv_2 ^2
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write 
mv_2^2 = mv^2

Therefore, the two final speeds of the balls are
v_2 = v
v_1 = -v_2 = -v

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0
3 years ago
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In short, and in general:

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<span>Credit Unions, and in particular smaller local credit unions, struggle to match the level of convenience (ATMs and branches) that many banks provide their customers, although many CUs are part of shared networks which enhance the breadth of delivery channels available to their membersSome Credit Unions are limited in their product offeringsOne must qualify for membership <span>One must pay a membership fee to join. hope this helps!

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7 0
3 years ago
A projectile is fired with an initial speed of 150 m/s and angle of elevation 60°. (Recall g ≈ 9.8 m/s2. Round your answers to t
natka813 [3]

Answer:

1988 m

Explanation:

Range of a projectile

R = U²sin2∅/g...................... Equation 1

Where R = Range, U = Initial velocity, g = acceleration due to gravity, ∅ = Angle of projection.

Given: U = 150 m/s, ∅ = 60°

Constant: g = 9.8 m/s².

Substitute these values into equation 1

R = 150²(sin60°)/9.8

R = 22500(0.866)/9.8

R = 1988 m

Hence the Range of the projectile is 1988 m

4 0
3 years ago
As a positively charged object moves toward a negatively charged object, their potential energy increases. As a positively charg
Zielflug [23.3K]

Answer:

As a positively charged object moves toward another positively charged object, their potential energy increases. True

As a negatively charged object moves toward another negatively charged object, their potential energy increases True

Explanation:

When two like charges move towards each other, there is increase in potential energy because the motion is in opposition to the direction if the field. Hence when two negative or two positive charges move towards each other, there is an increase in potential energy.

However, potential energy is inversely related to the distance of separation of the charges.

4 0
3 years ago
The force of repulsion between to like charge table tennis balls is 8.2 X 10 ^-7 N if the charge on the two objects a 6.7 X10^-9
Svetllana [295]

Answer:

0.702 m

Explanation:

The magnitude of the electrostatic force between two charged objects is given by Coulomb's Law:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the objects

And the force is:

- Attractive if the two charges have opposite signs (+-)

- Repulsive if the two charges have same sign (++ or --)

In this problem we have:

F=8.2\cdot 10^{-7}N is the force between the two balls

q_1 = q_2 = 6.7\cdot 10^{-9}C is the charge on each ball

Solving for r, we find the separation between the balls:

r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(9\cdot 10^9)(6.7\cdot 10^{-9})^2}{8.2\cdot 10^{-7}}}=0.702 m

5 0
3 years ago
Read 2 more answers
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