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3 years ago

Acceleration vs. Mass

Physics

1 answer:

Tamiku [17]3 years ago

4 0

1st. Answer:as the mass of the cart increases, the acceleration the cart decreases.

2nd. An inverse relationship

Explanation:

Edge 2020 yu got my word

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A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, a

anastassius [24]

Answer:

Miller Indices are [2, 4, 3]

Solution:

As per the question:

Lattice Constant, C = $4.83 \AA$

Intercepts along the three axes:

$\bar{x} = 9.66 \AA$

$\bar{x} = 19.32 \AA$

$\bar{x} = 14.49 \AA$

Now,

Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.

Now, for the Miller Indices along the three axes:

a = $\frac{1}{9.66}$

b = $\frac{1}{19.32}$

c = $\frac{1}{14.49}$

To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:

a' = $\frac{\frac{1}{9.66}}{\frac{1}{4.83}} = \frac{1}{2}$

b' = $\frac{\frac{1}{19.32}}{\frac{1}{4.83}} = \frac{1}{4}$

c' = $\frac{\frac{1}{14.49}}{\frac{1}{4.83}} = \frac{1}{3}$

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3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time

Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

4 0

3 years ago