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3 years ago

how many times does the kinectic energy of a car increase when traveling 60 mph as opposed to traveling 30 mph

1 answer:

3 0

The car at 60 kph has 9 times more kinetic energy than the car traveling at 20 kph. This assumes that both cars have the same mass. Kinetic energy depends on the square of thee speed so if one car is going 3 times faster, its kinetic energy will be 3^2 ( = 9 ) greater. The car going at 60 kph will have 4 times the KE of the car going at 30 kph ( again assuming that the cars have the same mass.)

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A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti

vagabundo [1.1K]

Answer:

t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

$d = v. t + \dfrac{v^2}{2a}$

$t = \dfrac{d}{v} -\dfrac{v}{2a}$

t is the reaction time

$t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}$

t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

3 0

3 years ago

The picture below shows the setup for an experiment involving a 1000 mL beaker, sitting on a burner, filled with 500 mL of water

Vlad1618 [11]

Answer:

bye

Explanation:

bye

4 0

3 years ago

True or False: As the mass of an object increases, the gravitational pull increases

Lunna [17]

Answer: The answer is obviously True!!!!

Explanation: The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them. This means that the force of gravity increases with mass, but decreases with increasing distance between objects. ... However, the exponent on the mass terms is one.

7 0

3 years ago

A compact disc (CD) is read from the bottom by a semiconductor laser beam with a wavelength of 790 nm that passes through a plas

Ad libitum [116K]

Answer:

The value is $t = 110 nm$

Explanation:

From the question we are told that

The wavelength of the beam is $\lambda = 790 \ nm = 790 *10^{-9} \ m$

The refractive index is $n_r = 1.80$

Generally from the condition for destructive interference the depth of the pit is mathematically represented as

$t = [ m + \frac{1}{2} ] * \frac{\lambda}{n_r} * \frac{1}{2}$

Here m which is the order of the fringe is zero because both beams cancel out

$t = [ 0 + \frac{1}{2} ] * \frac{790 *10^{-9}}{ 1.80} * \frac{1}{2}$

=> $t = 110 *10^{-9} \ m$

=> $t = 110 nm$

7 0

3 years ago

A projectile thrown from a point P moves in such a way that its distance from P is always increasing. Find the maximum angle abo

Anna71 [15]

Answer

70.52°

Explanation

The distance between projectile's position and it's starting point at any time is given by the relation

r² = x² + y²

where x = horizontal distance covered and y = vertical distance covered

According to projectile motion the horizontal displacement is given by

x = v(x)t = v cos(θ) t

Also the vertical component is given by

y = v(y) t - 0.5gt² = v sin(θ) t - 0.5gt²

Substituting the x and y values into the r-equation yields,

r² = (v cos(θ) t)² + (v sin(θ) t - 0.5gt²)²

r² = v²(cos²(θ))t² + v²(sin²(θ))t² – (vg sin(θ))t³+ 0.25 g²(t^4)

r² = v²t² (cos²(θ)+ sin²(θ)) – (vg sin(θ))t³ + 0.25 g²(t^4)

r² = v²t² – (vg sin(θ))t³ + 0.25 g²(t^4)

Differentiate r with respect to t

r(dr/dt) = 2v²t - 3vg sin(θ)t² + g²t³

At maximum angle the projectile could have been thrown above the horizontal, dr/dt = 0

2v²t - 3vg sin(θ)t² + g²t³ = 0

Divide through by t

2v² - 3vg sin(θ)t + g²t² = 0

g²t² - 3vg sin(θ)t + 2v² = 0

This can be solved using the general law for quadratic equations

(-b ± √(b² - 4ac))2a

a = g², b = -3vg sin(θ) c = 2v²

t = ((3vg sin(θ)) ± √(9v²g²sin²(θ) - 8g²v²))/2g²

This equation makes sense when the value under the square root is positive, that is, the square root exists.

9v²g²sin²(θ) - 8g²v² > 0

9sin²(θ) - 8 > 0

Meaning sin²(θ) = 8/9

Sin θ = (2√2)/3

θ = 70.52°

QED!!!

6 0

3 years ago