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Dovator [93]
3 years ago
13

A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

he speed of the train is 75 km/h how long does it take the car to pass the train,and how far will the car have traveled in this time?
Physics
1 answer:
victus00 [196]3 years ago
7 0

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

V=\dfrac{x}{t}

Where x is the distance, t is the time and v is the speed. using the data that we have:

V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km

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xperiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are gi
creativ13 [48]

Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201)\neq0

d) 2+2+(-4)=0

e) 2+2+(-2)\neq0

f) 2+2+(-3) \neq0; 2+(-2)+3\neq0

g) 2+2+(-5)\neq0; 2+(-2)+5\neq0

h)200 + 200 +(-5)\neq0; 200+(-200)+5\neq0

6 0
3 years ago
What is the value of the activation energy of the uncatalyzed reaction in reverse?
alexgriva [62]
It would be: Activation Energy = 300 KJ

Hope this helps!
5 0
3 years ago
A projectile is fired at time t= 0.0 s from point 0 at the edge of a cliff, with initial velocity components of Vox = 30 m/s and
BARSIC [14]

Answer:

At t = 15.0 s the magnitude of the velocity is 58.31 m/s

Explanation:

The given parameters are;

V₀ₓ = 30 m/s

V_{0y} = 100 m/s

The time of flight of the projectile = 25 s

For projectile motion;

Vₓ = V₀ × cos(θ₀)

The magnitude of the velocity V = √(V₀ₓ² + V_{0y}²)

We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s

cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109

θ₀ = cos⁻¹(3/√109) = 73.3°

The components of the velocity after time t is given by the relations;

Vₓ = V₀ × cos(θ₀) = 30 m/s

V_y =  V₀ × sin(θ₀) - g×t

When V_y = 0, we have;

0 =  V₀ × sin(θ₀) - g×t

g×t  =  V₀ × sin(θ₀)  = 10·√109×0.958 = 100 m/s

t = 100/g = 100/10 = 10 s

The time to reach maximum height = 10 s

At 15.0 seconds, we have;

V_y =  V₀ × sin(θ₀) - g×t = 10·√109×0.958  - 10×15 = -50 m/s

Therefore, the projectile is returning at 50 m/s

The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.

5 0
3 years ago
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s344n2d4d5 [400]

Answer:

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3 0
3 years ago
A ray of light is moving from a material having a high indexof refraction into a material with a lower index of refraction.
luda_lava [24]

(a) Away from the normal

We can find the direction of bending of the ray of light by using Snell's equation:

n_1 sin \theta_1 = n_2 sin \theta_2

where we have:

n1, n2: index of refraction of the first and second medium

\theta_1, \theta_2; angle that the incident and the refracted ray form with the normal to the surface

Here, the light ray moves from a material with high index of refraction to a material with lower index, so we have

n_1 > n_2

Re-arranging Snell's law we find

sin \theta_2 = \frac{n_1}{n_2} sin \theta_1

since we have

\frac{n_1}{n_2}>1

this implies

sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the wavelength of the light in a vacuum

n is the refractive index

The equation can be rewritten as

\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2

and again it can be rewritten as

\lambda_2 = \frac{n_1}{n_2} \lambda_1

where

\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1

Therefore, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency remains the same

Wavelength and speed of a light ray depend on the medium in which the wave is travelling through, however the frequency does not depend on that, so it remains the same in the two mediums.

8 0
3 years ago
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