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3 years ago

If an object triples its velocity, how does this effect its KE?

1 answer:

3 0

If an object triples its velocity then the kinetic energy become 9 times the initial kinetic energy. i hope this helped!

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A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal

VladimirAG [237]

Answer:

1.53 s

Explanation:

Initially vertical component of velocity of the ball, uy = 7.5 m/s

Net displacement is vertical direction is zero, Δy =0

Use second equation of motion:

Δy = uy t + 0.5 a t²

Here, acceleration a = -g (g =9.8 m/s²)

Substitute all the values and solve for g

0 = 7.5 t -0.5 (9.8)t²

7.5 t = 4.9 t²

t = 1.53 s

8 0

3 years ago

Read 2 more answers

A girl is bouncing on a trampoline where is her gravitational potential energy a maximum and where is her kinetic energy maximum

Stella [2.4K]

Answer:

When you jump down, your kinetic is converted to potential energy of the stretched trampoline. The trampoline's potential energy is converted into kinetic energy, which is transferred to you, making you bounce up. At the top of your jump, all your kinetic energy has been converted into potential energy. Right before you hit the trampoline, all of your potential energy has been converted back into kinetic energy. As you jump up and down your kinetic energy increases and decrease.

7 0

3 years ago

Describe how charge is transferred from the ruler to the metal rod.

spin [16.1K]

When the ruler is broughı near the inetal knob, it repels electrons in the metal. Electrons move away froni the ruler and down the metal rod. The knob now has a positive charge. The thin pieces of metal foil at the bottom of the metal rod now have a negative charge.

3 0

3 years ago

Read 2 more answers

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

$v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)$

v₂ = 6.4 m/s

The final speed of the leading car is given by the following formula:

$v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)$

v₁ = 3.5 m/s

4 0

3 years ago

Help please It’s kinda urgent

user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

$(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]$

The negative sign of acceleration means that the ship slows down its velocity in order to land.

4 0

2 years ago