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Illusion [34]
4 years ago
10

How do Newton's laws of motion apply to the motion of an animal, such as a cat that is running?

Physics
2 answers:
Bad White [126]4 years ago
8 0
An object in motion will still be in motion and an object at rest will stay at rest unless being acted on by an unbalanced force
Zolol [24]4 years ago
3 0
According to newton's first law of motion a stationary or moving object will be in that state forever until an external force is exerted
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Select the correct answer. In a given chemical reaction, the energy of the products is greater than the energy of the reactants.
ioda
B. Energy is released in the reaction... just looked it up!
3 0
3 years ago
It takes 3 minutes to make toast in a 1500 watt toaster. Calculate how much work is done by the toaster.
REY [17]

Answer: 2.7 x 10^5 joules

Explanation:

Given that:

Time taken = 3 minutes

convert time in minutes to seconds

(Since 1 minute = 60 seconds

3 minutes = 3 x 60 = 180 seconds)

Power of toaster = 1500 watt

Work done by the toaster = ?

Recall that power is the rate of work done per unit time

i.e Power = work/time

work = Power x Time

Work = 1500 watt x 180 seconds

Work = 270000J

Place the result in standard form

270000J = 2.7 x 10^5J

Thus, 2.7 x 10^5 joules of work is done by the toaster.

3 0
3 years ago
By method of dimension show that the following equation are homogenous.
il63 [147K]

Answer:

Proof in explanataion

Explanation:

The basic dimensions are as follows:

MASS = M

LENGTH = L

TIME = T

i)

Given equation is:

H = \frac{u^2Sin^2\phi}{2g}

where,

H = height (meters)

u = speed (m/s)

g = acceleration due to gravity (m/s²)

Sin Ф = constant (no unit)

So there dimensions will be:

H = [L]

u = [LT⁻¹]

g = [LT⁻²]

Sin Ф = no dimension

Therefore,

[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]

<u>[L] = [L]</u>

Hence, the equation is proven to be homogenous.

ii)

F = \frac{Gm_1m_2}{r^2}\\\\

where,

F = Force = Newton = kg.m/s² = [MLT⁻²]

G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²

G = [M⁻¹L³T⁻²]

m₁ = m₂ = mass = kg = [M]

r = distance = m = [L]

Therefore,

[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\

<u>[MLT⁻²] = [MLT⁻²]</u>

Hence, the equation is proven to be homogenous.

8 0
3 years ago
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