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2 years ago

Which has more kinetic energy, a 4.0 kg bowling ball moving at 1.0 m/s or a 1.0 kg

Physics

1 answer:

jolli1 [7]2 years ago

5 0

Answer:

The kinetic energy of bocce ball is more.

Explanation:

Given that,

Mass of a bowling ball, m₁ = 4 kg

Speed of the bowling ball, v₁ = 1 m/s

Mass of bocce ball, m₂ = 1 kg

Speed of bocce ball, v₂ = 4 m/s

We need to say which has more kinetic energy.

The kinetic energy of an object is given by :

$E=\dfrac{1}{2}mv^2$

Kinetic energy of the bowling ball,

$E_1=\dfrac{1}{2}m_1v_1^2\\\\E_1=\dfrac{1}{2}\times 4\times (1)^2\\\\E_1=2\ J$

The kinetic energy of the bocce ball,

$E_2=\dfrac{1}{2}m_2v_2^2\\\\E_2=\dfrac{1}{2}\times 1\times (4)^2\\\\E_2=8\ J$

So, the kinetic energy of bocce ball is more than that of bowling ball.

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during conversion water to ice,forces of attraction between molecules. OPTIONS:1.)Decreases 2.)Increases 3.)Does not change 4.)C

In-s [12.5K]

because water is loosely packed but when it is cold it becomes closely packed in order to form ice and thus the force attraction between them also increase.

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3 years ago

A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.45 m/s. knowing that

lakkis [162]

When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:

H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s

t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s

Total time = 0.125 s + 4.495 s = 4.62 seconds

8 0

3 years ago

At 2 P.M., ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast i

labwork [276]

Answer:

The distance between the ships changing at 6PM is 21.29Km/h

Explanation:

Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h

Given

dx/dt= 35

dy/dt= 25

dv/dt= ???? at t= 6PM - 2PM= 4

Therefore t=4

We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction

So, we use:

$D^2 = (150 - x)^2 + y^2$ ;

$D^2 = (150 - 140)^2 + y^2$

But ship B travels at t=4, at 4.25 =100 in the y-direction

so, let's use the equation:

$D^2 = 10^2 + 100^2$

$= D= sqrt*(10 + 100)$

Lets use 2DD' = 2xx' + 2yy'

Differentiating with respect to t we have:

D•d(D)/dt = -(10)•dx/dt + 100•dy/dt

=100.5 d(D)/dt = (-10)•35 + (100)•25

When t=4, we have x=(140-150) =10 and y=100

$= D = sqrt*(10^2 + 100^2)$

=100.5

= 100.5 dD/dt = 10.35 +100.25

= dD/dt = 21.29km/h

7 0

2 years ago

A car travels in a flat circle of radius R. At a certain instant the velocity of the car is 24 m/s west and the total accelerati

lisov135 [29]

Answer with Explanation:

We are given that

Velocity,v=24 m/s west

Acceleration,a= $2.5m/s^2$

$\theta=53^{\circ}$ (N of W)

Horizontal component of acceleration=Tangential acceleration

Tangential acceleration, $a_x=acos\theta=2.5cos53=1.505 m/s^2$

Radial acceleration=Vertical acceleration= $a_y=asin\theta$

$a_y=2.5sin53=1.998 m/s^2$

Radial acceleration, $a_y=\frac{v^2}{R}$

$1.998=\frac{(24)^2}{R}$

$R=\frac{(24)^2}{1.998}$

$R=288.29 m$

Time, $t=\frac{2\pi R}{velocity}$

$t=\frac{2\pi(288.29)}{24}$

$t=75.47 s$

8 0

3 years ago

A pendulum is swinging back and forth with a period of 2.0 seconds here on Earth. This pendulum is then brought to the Moon, whe

leonid [27]

Answer: The period of the pendulum will be bigger than in Earth.

Explanation:

The period of a pendulum is:

T = 2*pi*√(L/g)

where pi = 3.14

L is the length of the pendulum and g is the gravitational acceleration.

you can see that g is in the denominator, so if g is smaller, then the end number woll be bigger (because in the moon we are dividing by a smaller number)

This means that in the moon the period of the pendulum is bigger than in the Earth.

5 0

3 years ago