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2 years ago

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A soccer ball is kicked 8 m north. Then a teammate kicks it 6 m east into the goal. What is the soccer ball’s displacement from

where it was first kicked?
10 m northeast
14 m northeast
100 m northeast
4 m southeast

2 answers:

Lynna [10]2 years ago

7 0

Answer:

Displacement=10 meters

Explanation:

The displacement is the direct path from the first contact with the ball and the last spot it was seen in. The ball was kicked north 8 meters, meaning it was kicked vertically upward, and then it was kicked east 6 meters, or to the right. If we were to draw the displacement we would see that this is a right triangle. That means we can use the Pythagorean Theorem:

$a^2+b^2=c^2\\8^2+6^2=c^2\\64+36=c^2\\c^2=100\\\sqrt{c} =\sqrt{100} \\c=10$

Boom.

alex41 [277]2 years ago

4 0

Answer:

Your cool epic and fast

Explanation:

I lik amige ssahashda

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A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at

docker41 [41]

Answer:

$y = 4.36$

Explanation:

Let the height of the ladder be L

$L = 10$

Also:

Let $x = distance\ from\ the\ base\ of\ the\ ladder$

Let $y = height\ of\ the\ base\ of\ the\ ladder$

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

$L^2 = x^2 + y^2$ --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

$x = 9$

Substitute 9 for x and 10 for L in $L^2 = x^2 + y^2$

$10^2 = 9^2 + y^2$

$100 = 81 + y^2$

Make $y^2$ the subject

$y^2 = 100 - 81$

$y^2 = 19$

Make y the subject

$y = \sqrt{19$

$y = 4.36$

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0

3 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

3 0

3 years ago

Light is a form of what

mart [117]

Light is a form of energy.
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8 0

3 years ago

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

3 years ago

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

3 years ago