Ask question

Ask question

All categories

Liono4ka [1.6K]

3 years ago

13

Light traveling through air strikes the surface of the four different materials shown. Which material reflects light but does no

t refract it? 1. Diamond 2. water being poured 3. triangular glass prism 4. 2 silver spoons

2 answers:

AysviL [449]3 years ago

7 0

Answer:

The answer is 4. 2 silver spoons

Explanation:

Hello! Let's solve this!

Many materials refract light in addition to reflecting it. This means that the light passes through the object at a certain angle.

In this case, of the given options, the correct option is 4. 2 silver spoons, since it is the only material that is not transparent of the dice.

We conclude that the answer is 4. 2 silver spoons

Elza [17]3 years ago

4 0

It would be{ 4) 2 silver spoons}<span>
that the material reflects light but does not refract it</span>

You might be interested in

A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisher

Anna71 [15]

here we can say that there is no external force on fisherman and dock

so here we will use momentum conservation theory

As per momentum conservation

initial momentum of fisherman + boat = final momentum of fisherman + boat

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we will have

$83(3.1) + 139(0) = 83 v + 139 v$

$257.3 = 222v$

$v = 1.16 m/s$

so the speed of boat and fisherman will be 1.16 m/s

3 0

3 years ago

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

$d = v*t$

Where:

v: is the tangential speed of the disk

t: is the time = 30 s

The tangential speed can be found as follows:

$v = \omega*r$

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

$v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s$

Now, the distance traveled by the disk is:

$d = v*t = 5.24 m/s*30 s = 157.2 m$

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0

2 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

7 0

3 years ago

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder

kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0

3 years ago

Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t

Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N

Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.

8 0

3 years ago