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3 years ago

How do I know what kinematic equations to use when solving a question?

Physics

1 answer:

olganol [36]3 years ago

5 0

Explanation:

There are 5 kinematic equations, and 5 variables.

Each question will give you 3 variables and ask you to solve for a fourth.

To determine which equation to use, look at which variable is <em>not</em> included in the problem.

For example, if the question does not include time, then you need to use a kinematic equation that does not have t in it. That would be:

v² = v₀² + 2aΔx

Or, if the question does not include the final velocity, then you need a kinematic equation that does not have v in it. That would be:

Δx = v₀ t + ½ at²

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valentinak56 [21]

Answer:

A dump truck going 70 mph

Explanation:

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Find the KE of a ball of mass 500 g moving with a velocity of 4 m/s . Plz fast i have just 5 mins

creativ13 [48]

Answer:

4000 J

Explanation:

KE = (1/2)(m)(v^2)

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KE = 4000 J

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3 years ago

A shell is shot with an initial velocity v with arrow0 of 18 m/s, at an angle of θ0 = 60° with the horizontal. At the top of the

BARSIC [14]

Answer:

D = 43 m

Explanation:

given,

initial velocity = 18 m/s

angle θ = 60°

total horizontal distance covered by the shell is

$R = \dfrac{v_0^2sin 2\theta}{g}$

applying conservation of momentum in horizontal direction

m v₀ cos θ = m₁v₁ + m₂ v₂

m v₀ cos θ = 0.5 m v₂

v₂ = 2 v₀ cos θ.

distance covered by the shell from point of explosion

R' = v t

= $(2 v_0 cos \theta) (\dfrac{v_0^2sin \theta}{g})$

= $(2 \dfrac{v_0^2cos \theta sin \theta}{g})$

= $\dfrac{v_0^2sin 2\theta}{g}$

= R

total distance traveled by the shell is

D = $\dfrac{R}{2}+R'$

= 1.5 R

= $1.5\dfrac{v_0^2sin 2\theta}{g}$

D = $1.5\dfrac{18^2sin 2\times 60}{9.81}$

= 42.9 ≅ 43 m

D = 43 m

3 0

3 years ago

The age of a piece of wood from an archeological site is to be determined using the Carbon-14 method. The activity of the sample

Bezzdna [24]

Answer:

4589.05 year

Explanation:

The relation of activity is given by $A=A_0e^{\lambda t}$ we have given $\frac{A}{A_0}=0.574$

Half life $T=5730\ years$

We know that half life $T=\frac{0.693}{\lambda }$

$\lambda =\frac{0.693}{5730}=1.2094\times 10^{-4}\ per\ year$

So $0.574=e^{-1.2094\times 10^{-4}t}$

$e^{1.2094\times 10^{-4}t}=\frac{1}{0.574}=1.742$

${1.2094\times 10^{-4}t}=ln1.742=0.555$

$t=\frac{0.555}{1.2094\times 10_{-4}}=4589.05\ year$

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3 years ago

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Answer:

The correct answer should be

A. 20 Joules

Explanation:

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