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MatroZZZ [7]
3 years ago
11

Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?

Physics
1 answer:
aliina [53]3 years ago
7 0

Answer:

15 miles

Explanation:

6 miles per hour

2 1/2 hours

6 x 2 = 12

6 x 1/2 = 3

12 + 3 = 15

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Which of the following is not a valid conversion factor?
kykrilka [37]
The third option is wrong
8 0
3 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
True or False? 13. All living things are made of cells 14. All cells have DNA within their nucleus 15. The cell is the basic uni
Leni [432]

Answer:

1: True

2: False

3: True

4: False

5: True

3 0
3 years ago
The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency? Hz(b) What is its wavelength in glas
vekshin1
<h2>Answers:</h2>

The speed of a wave is given by:

v=f.\lambda  (1)

Where f is the frequency and  \lambda the wavelength.

In the case of light, its speed is:

c=f.\lambda (2)

On the other hand, the described situation is known as Refraction,   a phenomenon in which the light changes its direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index n is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum (c=3(10)^{8}m/s) and the speed of light v in the second medium:

n=\frac{c}{v} (3)

In addition, as the light changes its direction, its wavelength changes as well:

n=\frac{\lambda_{air}}{\lambda_{glass}} (4)

Knowing this, let's begin with the answers:

<h2>a) Frequency</h2>

From equation (2) we can find f:

f=\frac{c}{\lambda}  (5)

Knowing that 1nm=(10)^{-9}m:

f=\frac{3(10)^{8}m/s}{632.8(10)^{-9}m}  

f=4.74(10)^{14}Hz}     (6)   >>>Frequency of the helium-neon laser light

<h2>b) Wavelength in glass</h2>

We already know the wavelength of the light in air \lambda_{air} and the index of refraction of the glass.

So, we only have to find the wavelength in glass \lambda_{glass} from equation (4):

\lambda_{glass}=\frac{\lambda_{air}}{n}

\lambda_{glass}=\frac{632.8(10)^{-9}m}{1.48}

\lambda_{glass}=427(10)^{-9}m=427nm   (7)   >>>Wavelength of the helium-neon laser light in glass

<h2>c) Speed in glass</h2>

From equation (3) we can find the speed vof this light in glass:

v=\frac{c}{n}

v=\frac{3(10)^{8}m/s}{1.48}

v=2.027(10)^{8}m/s   (8)  >>>Speed of the helium-neon laser light in glass

7 0
3 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
4 years ago
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