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My name is Ann [436]
2 years ago
5

the deflection angle of the laser beam as it exits the prism is 22. 6º. If the prism had been made of glass instead of polystyre

ne plastic, what would the deflection angle have been?.
Physics
1 answer:
inessss [21]2 years ago
4 0
Thirty three degrees
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Una rueda que tiene 15 cm de radio, realiza 64 vueltas en 16 seg. Calcula: Periodo Frecuencia Velocidad angular Velocidad lineal
liraira [26]

Answer:

i) El período de la rueda es de 0,25 segundos.

ii) La frecuencia de la rueda es 4 Hertz

iii) La velocidad angular es aproximadamente 25.133

iv) La velocidad lineal es de aproximadamente 3,77 m / s

Explanation:

El radio de la rueda, r = 15 cm = 0,15 m

El número de vueltas que hace la rueda = 64 vueltas

El tiempo que tarda el volante en dar 64 vueltas = 16 segundos

i) El período = El tiempo que tarda la rueda en dar 1 vuelta

∴ El período de la rueda, T = 16 segundos/(64 vueltas) = 0,25 segundos

El período de la rueda, T = 0,25 segundos

ii) La frecuencia = El número de vueltas por segundo

∴ La frecuencia de la rueda, f = 64 vueltas /(16 segundos) = 4 Hertz

1 vuelta = 2 · π radianes

La frecuencia de la rueda, f = 4 Hertz

iii) Velocidad angular = La medida del ángulo girado por segundo

∴ La velocidad angular, ω = 64 × 2 × π/16 segundos ≈ 8 · π rad/segundos ≈ 25.133 rad/seg

La velocidad angular, ω ≈ 25.133

iv) La velocidad lineal, v = r × ω

∴ v = 0,15 m × 8 · π rad / segundos ≈ 3,77 m/s

La velocidad lineal, v ≈ 3.77 m/s

7 0
3 years ago
Help it’s multiple choice 11 through 15 please!
riadik2000 [5.3K]

1. • Here, force of gravity on the block = 20 N.

• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]

• The coefficient

u_{k} = 0.4

• Force of friction =

u_{k} \times  \: normal \:  \:  \: force \\  = 0.4 \times 20N \\  = 8N

• Hence, the force of sliding friction between the block and the ground is 8 N.

• So, it is option c. 8 N

2. The answer is option d. continue in the same direction with no change in speed.

We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.

3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.

Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.

So, 1 N = 1 Kg × 1 m/s^2.

4. It is d. not zero.

Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.

5. Force = 2 N

Acceleration of the object A = 2 m/s^2.

Acceleration of the object B = 1 m/s^2.

Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg

And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg

So, the mass of object B is greater than that of object A.

Hence, the answer is option c. Object B has more mass.

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0
2 years ago
What accurately describes what happens when water vapor condenses into dew in terms of energy
Andreyy89
The awnser is condensation
5 0
2 years ago
Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be
ANTONII [103]

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ -  2.5 x 10⁵  x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

5 0
3 years ago
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

#SPJ1

4 0
2 years ago
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