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3 years ago

You throw a baseball that has a mass of 0.425 kilograms. The ball leaves your hand with a speed

Physics

1 answer:

shutvik [7]3 years ago

6 0

Answer:

430.3125 J

Explanation:

Kinetic Energy (KE) is the energy type that a object gets due to its motion.

Therefore the baseball has a Kinetic Energy

$KE = \frac{1}{2} *mass*velocity^{2} \\=\frac{1}{2} *0.425*45^{2}\\=430.3125 J$

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Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.

marusya05 [52]

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

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2 years ago

how many years would it take to reach the star from earth, as measured by observers on the spacecraft

Assoli18 [71]

In other words, it would take Deep Space 1 more than 81,000 years to travel the 4.24 light-years between Earth and Proxima Centauri at its top speed of 56,000 km/h. In relation to human history, that would be more than 2,700 generations.

Nearly 40 trillion kilometers, or 4.4 light-years, separate us from Alpha Centauri. The NASA-Germany Helios probes, the fastest spacecraft to date to be launched into orbit, flew at a speed of 250,000 kilometers per hour. The probes would need 18,000 years to travel at such pace to arrive at the sun's nearest neighbor. The calculations reveal that it is almost impossible to reach the nearest star in a human lifetime, even with the most futuristic technologies.

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brainly.com/question/1302132

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3 0

2 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0

3 years ago

8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h

pantera1 [17]

(8) A car starting with a speed v skids to a stop over a distance d, which means the brakes apply an acceleration a such that

0² - v² = 2 a d → a = - v² / (2d)

Then the car comes to rest over a distance of

d = - v² / (2a)

Doubling the starting speed gives

- (2v)² / (2a) = - 4v² / (2a) = 4d

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration a such that

0² - (13.9 m/s)² = 2 a (15 m) → a ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance d such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) d → d ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass m such that

60 J = m g h

where g = 10 m/s² and h is the height it is lifted, 1.2 m. Solving for m gives

m = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0

3 years ago