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3 years ago

How do you calculate acceleration

Physics

1 answer:

Eduardwww [97]3 years ago

7 0

Answer: acceleration is equal to the change in velocity per unit time in seconds.

a= ∆v / t = vf - vi / t

Explanation: change in velocity or ∆v can be expressed as (vf - vi)

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Match the terms to the correct descriptions.

Ulleksa [173]

2. kinetic energy: due to it being transferred through collisions

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3 years ago

Read 2 more answers

A father pushes his child in a cart. The cart starts to move.

LuckyWell [14K]

The potential energy of the car when it let go is 20,000 J.

The speed of the car at the bottom of the ramp is 20 m/s.

The given parameters;

mass of the car, m = 100 kg
height of the car, h = 20 m

The potential energy of the car is calculated as follows;

P.E = mgh

P.E = 100 x 10 x 20

P.E = 20,000 J

The speed of the car at the bottom of the ramp is calculated as follows;

$K.E = P.E\\\\\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2 \times 10 \times 20} \\\\v = 20 \ m/s$

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3 years ago

Consider a car that travels between points A and B. The car's average speed can be greater than the magnitude of its average vel

ryzh [129]

Answer: A.True

Explanation: simply put the magnitude of the speed is a function of Distance while the magnitude of velocity is a function of Displacement. Displacement is the average Distance moved in different directions which will be smaller in magnitude compared to the Total Distance used the calculating the magnitude of speed.

4 0

4 years ago

Someone help please i need to finish this

ICE Princess25 [194]

Answer:

Fluorine

Explanation:

The more number of electrons on the outer shell makes the atom more reactive

8 0

3 years ago

X- rays with a wavelgnth of 0.0700 nm diffract from a crystal. Two adjacne angels of x-ray diffraciton are 45.6 and 21.0 degrees

zhenek [66]

Answer:

0.15 nm

Explanation:

d = Distance between the atomic planes

m = Order

$\theta_{m}$ = First angle = $45.6^{\circ}$

$\theta_{m+1}$ = Adjacent angle = $21^{\circ}$

$\lambda$ = Wavelength = 0.07 nm

From Bragg's relation we know

$2d\cos\theta_{m}=m\lambda$

$2d\cos45.6^{\circ}=m0.07$

$2d\cos\theta_{m+1}=(m+1)\lambda$

$2d\cos21^{\circ}=(m+1)0.07\\\Rightarrow 2d\cos21^{\circ}=m(0.07)+0.07$

$2d\cos21^{\circ}=2d\cos45.6^{\circ}+0.07\\\Rightarrow 2d(\cos21^{\circ}-\cos45.6^{\circ})=0.07\\\Rightarrow d=\dfrac{0.07}{2(\cos21^{\circ}-\cos45.6^{\circ})}\\\Rightarrow d=0.14962\ \text{nm}$

The distance between the atomic planes is 0.15 nm.

4 0

3 years ago