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3 years ago

12

5. Your friend tosses a ball into the air. The equation h = -5ť + 30t +10 models the height,

1 answer:

Naily [24]3 years ago

4 0

Answer:

6.3s.

Explanation:

The height h is given by:

$h(t)=-5t^2+30t+10$

Setting the height h = 0 and using the quadratic formula to solve for t:

$0=-5t^2+30t+10\\0=t^2-6t-2\\t_1=3-\sqrt{11}= -0.3s\\t_2=3+\sqrt{11}=6.3s$

t₁ is physicly meaningless.

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What do you mean by focus in science

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“a point at which rays of light, heat, or other radiation meet after being refracted or reflected.” Meaning multiple light rays or heat (and other forms of radiation) are all being refracted or reflecting to a certain point

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2 years ago

A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

so:

Mgh = $\frac{1}{2}IW^2 +\frac{1}{2}MV^2$

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = $\frac{1}{2}MR^2$

I = $\frac{1}{2}(5kg)(2m)^2$

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = $\frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2$

Replacing the data:

(5kg)(9.8)(0.3m) = $\frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2$

solving for V:

(5kg)(9.8)(0.3m) = $V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))$

V = 2 m/s

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3 years ago

The site of protein synthesis is

vovangra [49]

Ribosomes is your answer

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3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

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Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

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3 years ago

A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What

antiseptic1488 [7]

Answer: -4000 kg • m/s

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2 years ago