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1 year ago

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What horizontal velocity should be given to a ball hanging on a 4.9 m long thread so that it slides to the height of the hanger

(do not take into account the force of air resistance)

1 answer:

igomit [66]1 year ago

5 0

Answer:

13 u2 = 0 – (2 × –2 m/s2 × 6 m) = 24, u = 4.9 m

Explanation:

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Answer:
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Explanation:

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Is one type of igneous rock that has erupted from volcanoes or seafloor fissures, or freeze at shallow depths. this type of rock

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3 years ago

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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

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2 years ago

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