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3 years ago

How much kinetic energy does an 8 kg bowling ball have when it is thrown 9 m/s. down the alley

Physics

1 answer:

nika2105 [10]3 years ago

4 0

Answer:

324 J

Explanation:

KE= 1/2 m*v^2

1/2*8*9^2

324 joules (=^w^=) woof woof

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a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v

solniwko [45]

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

S = ut + ½ gt²

Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)

Adding these two equations,

x + (S - x) = U tₓ

S = U tₓ

200 = 50 tₓ

∴ tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s

6 0

3 years ago

Work done in taking charge from one point of a conductor to is another point is called

Yuliya22 [10]

Answer:

⁸

Explanation:

electric potential

I think so

6 0

3 years ago

2 Points

Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

$:\implies$ Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

$:\implies$ 50 N = 10 kg × Acceleration

$:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}$

$:\implies$ Acceleration of the Object = 5 m/s²

4 0

3 years ago

Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm

Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

= 9 x 10⁹ x [ q₁q₂ / r₁₂ + q₂q₃ / r₂₃ + q₁q₃ / r₁₃ ]

here r₁₂ , r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy = 9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08 + -200x10⁻¹⁸ / .04 + 200 x 10¹⁸ / .04 ]

= 9 x 10⁹ x - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0

3 years ago

Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl

Vera_Pavlovna [14]

Answer:

A, B, C, E

Explanation:

now gimme thanks please

4 0

3 years ago

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