Tom puts a metallic cover over his car's windshield after parking. How does this control

Is the average speed of a vehicles vector or a scalar quantity? A) vectorB) scalar

As the speed is a scalar quantity as it has the only magnitude in it. Therefore, the average speed is also stated as a scalar quantity.

Hence, the correct answer is (B)

3 0

1 year ago

If two proton approach each other with speed of 400m\s and 450m\s respectively what is the total momentum of the two particle sy

lilavasa [31]

Answer: $83.65\times 10^{-27}\ kg.m/s$

Explanation:

Given

The velocities of the protons are 400 m/s and 450 m/s

Mass of the Proton is $m=1.673\times 10^{-27}\ kg$

Since they travel in the opposite direction, therefore, their momentum cancels out each other.

Momentum is the product of mass and velocity

$\therefore m(450-400)\\\Rightarrow 1.673\times 10^{-27}\times 50\\\Rightarrow 83.65\times 10^{-27}\ kg.m/s$

The direction of momentum is in the direction of proton having 450 m/s velocity.

7 0

2 years ago

True or False: Cooler objects always have more thermal energy than warmer objects.

tensa zangetsu [6.8K]

False

Because you can see using a thermal camera I guess

3 0

3 years ago

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li

Fofino [41]

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement

t₁ : Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁: Stan acceleration

d₂: car displacement

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂: Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ = 1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1 = equation 2

1.55*t₁² = 2.495* t₂² , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² = 2.495* t₂²

1.55* (t₂² +2t₂+1) = 2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0 Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁ t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂

vf₂ =0+4.99* 3.87

vf₂ = 19.31 m/s : Kathy's final speed

8 0

3 years ago

Any help is appreciated please

stiv31 [10]

Answer:

C. It speeds up, and the angle increases

Explanation:

We can answer by using the Snell's law:

$n_i sin \theta_i = n_r sin \theta_r$

where

$n_i, n_r$ are the refractive index of the first and second medium

$\theta_i$ is the angle of incidence (measured between the incident ray and the normal to the surface)

$\theta_r$ is the angle of refraction (measured between the refracted ray and the normal to the surface)

In this problem, light moves into a medium that has lower index of refraction, so

$n_r < n_i$

We can rewrite Snell's law as

$sin \theta_r =\frac{n_i}{n_r}sin \theta_i$

and since

$\frac{n_i}{n_r}>1$

this means that

$sin \theta_r > sin \theta_i$

which implies

$\theta_r > \theta_i$

so, the angle increases.

Also, the speed of light in a medium is given by

$v=\frac{c}{n}$

where c is the speed of light and v the refractive index: we see that the speed is inversely proportional to n, therefore the lower the index of refraction, the higher the speed. So, in this problem, the light will speed up, since it moves into a medium with lower index of refraction.

4 0

3 years ago