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tamaranim1 [39]

3 years ago

7

6. A pitcher throws a ball toward home plate 18.39 meters away. If the ball is traveling at a constant 40.0 m/s, how long does i

t take to get there? *

1 answer:

Kruka [31]3 years ago

3 0

18.39 I don’t know this

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A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal

ss7ja [257]

Answer:

$v_{o}=141.51m/s$

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

$x=1420m\\g=-9.8m/s^{2}$

From the equation of x-position we know that

$x=v_{ox}t=v_{o}cos(35)t$

Solving for $v_{o}$

$v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)}$ (1)

Now, if we analyze the equation of y-position we got

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0

$0=v_{o}sin(35)t+\frac{1}{2}gt^{2}$

Knowing the equation for $v_{o}$ in (1)

$0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}$

$\frac{1}{2}(9.8)t^{2}=1420tan(35)$

Solving for t

$t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s$

Now, we can solve (1)

$v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s$

6 0

3 years ago

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The downward acceleration of a falling body on Earth is 9.81m/s2. On the moon the same quantity is 1.62m/s2. An astronaut in a s

Bas_tet [7]

The astronaut's mass doesn't change. It's the same wherever he goes,
because it doesn't depend on what else is around him.

His weight depends on what else is near him, so it changes, depending
on where he is.

   Weight = (mass) x (gravity)

On Earth, Weight = (145 kg) x (9.81 m/s²) = 1,422.5 newtons.
   (about 320 pounds)

On the moon, Weight = (145 kg) x (1.62 m/s²) = 234.9 newtons.
   (about 53 pounds)

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3 years ago

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Answer:

poda patiiiiiiiiii tadiya kundiii

5 0

3 years ago

point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is

Sergio [31]

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, $\rho_{w} = 1000\ kg//m^{3}$

Acceleration due to gravity, g = $9.8\ m/s^{2}$

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = $\frac{H + D}{2}$

h = $\frac{3 + 2.5}{2} = 2.75\ m$

Volume of water in the pool, V = $\pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}$

Mass of water, $m_{w} = \frac{\rho_{w}}{V}$

$m_{w} = \frac{1000}{311.02} = 3.215\ kg$

Work done is given by the potential energy of the water as:

$W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J$

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done, $W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J$

7 0

3 years ago

Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they h

omeli [17]

Answer:

Explanation:

Considering non - relativistic approach : ----

Speed of electron = 1 % of speed of light

= .01 x 3 x 10⁸ m /s

= 3 x 10⁶ m /s

Kinetic energy of electron = 1/2 m v²

= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²

= 40.95 x 10⁻¹⁹ J

Kinetic energy in electron comes from lose of electrical energy equal to

Ve where V is potential difference under which electron is accelerated and e is electronic charge .

V x e = kinetic energy of electron

V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹

V = 25.6 Volt .

6 0

3 years ago