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2 years ago

Which type of mirror causes beams of light to spread apart?.

Physics

1 answer:

DIA [1.3K]2 years ago

7 0

Answer: convex mirror

Explanation: <em>A convex mirror, on the other hand, has a surface that curves outward, like the back of a spoon. Convex mirrors cause light waves to spread out, or diverge. These two types of mirrors form images that are different from the images that are formed by plane mirrors.</em>

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Earth's inner core is:a dense ball of solid metal

hram777 [196]

What is one effect of steroid abuse in professional sports?
AThey cause athletes to stop training hard incorrect answer
BThey allow some competitors to gain an unfair advantage incorrect answer
CThey cause athletes to stop competing to win incorrect answer
DThey cause athletes to become selfish

7 0

3 years ago

You're driving down the highway late one night at 21 m/s when a deer steps onto the road 35 m in front of you. Your reaction tim

andreev551 [17]

Speed with which initially car is moving is 21 m/s

Reaction time = 0.50 s

distance traveled in the reaction time d = v t

d = 21 * 0.50 = 10.5 m

deceleration after this time = -10 m/s^2

now the distance traveled by the car after applying bakes

$v_f^2 - v_i^2 = 2a d$

$0 - 21^2 = 2(-10)d$

$d = 22.05 m$

so total distance moved before it stop

d = 22.05 + 10.5 = 32.55 m

so the distance from deer is 35 - 32.55 = 2.45 m

now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop

so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m

again by kinematics

$v_f^2 - v_i^2 = 2 ad$

$0 - v^2 = 2(-10)(24.5)$

$v = 22.1 m/s$

so maximum speed would be 22.1 m/s

5 0

3 years ago

Since the moon has less mass than the earth what happens to objects on the moon

postnew [5]

When you drop an object on the moon, it falls to the ground.
But it only falls about 1/6 as fast as it falls on Earth.

4 0

3 years ago

An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring

lozanna [386]

Answer:

0.6 m

Explanation:

When a spring is compressed it stores potential energy. This energy is:

Ep = 1/2 * k * x^2

Being x the distance it compressed/stretched.

When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.

The potential energy of the ice cube is:

Ep = m * g * h

This is vertical height and is related to the distance up the slope by:

sin(a) = h/d

h = sin(a) * d

Replacing:

Ep = m * g * sin(a) * d

Equating both potential energies:

1/2 * k * x^2 = m * g * sin(a) * d

d = (1/2 * k * x^2) / (m * g * sin(a))

d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m

8 0

3 years ago

There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor

VMariaS [17]

Answer:

F = 7.68 10¹¹ N, θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

cos 45 = F₂₄ₓ / F₂₄

sin 45 = F_{24y) / F₂₄

F₂₄ₓ = F₂₄ cos 45

F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

Fₓ = -F₂₁ + F₂₄ₓ

Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis

F_y = - F₂₃ + F_{24y}

F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

F₂₁ = F₂₃

Let's use Coulomb's law

F₂₁ = k q₁ q₂ / r₁₂²

the distance between the two charges is

r = a

F₂₁ = k q² / a²

we calculate F₂₄

F₂₄ = k q₂ q₄ / r₂₄²

the distance is

r² = a² + a²

r² = 2 a²

we substitute

F₂₄ = k q² / 2 a²

we substitute in the components of the forces

Fx = $- k \frac{q^2}{a^2} + k \frac{q^2}{2 a^2} \ cos 45$

Fx = $k \frac{q^2}{a^2}$ ( -1 + ½ cos 45)

F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)

We calculate

F₀ = 9 10⁹ 4.25² / 0.440²

F₀ = 8.40 10¹¹ N

Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

Fₓ = -5.43 10¹¹ N

remember cos 45 = sin 45

F_y = - 5.43 10¹¹ N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

F = -5.43 10¹¹ (i + j) N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

F = $\sqrt{F_x^2 + F_y^2}$

F = 5.43 10¹¹ √2

F = 7.68 10¹¹ N

in angle is

θ = 45º

7 0

3 years ago