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3 years ago

What is the volume of a 6-sided die a cube with each side measuring 1.6 cm across

Physics

1 answer:

Savatey [412]3 years ago

8 0

4.096. You just have to multiply 1.6 three times. V=s^3

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To Play Ice Hockey each player needs

Sholpan [36]

Answer:

If one plays ice hockey he should must have necessary equipment to support his sport. These equipment include Ice Skates,Helmet with Cage and Mouth-guard:, Hockey stick, Hockey pants,Hockey gloves, shoulder pads, elbow pads, Shin Guard:Neck guard and Jockstrap (men) or Pelvic protector (women)

Explanation:

7 0

3 years ago

Read 2 more answers

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance

xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}$

$\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}$

$v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}$

$v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}$

$v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}$

v₂ = 7.6 x 10⁴ m/s

3 0

3 years ago

011 10.0 points

Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

$v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s$

and the time taken to cover the horizontal distance d is

$t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s$

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

$y=u_y t - \frac{1}{2}gt^2$

where

$u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s$ is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

$y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m$

The height of the crossbar is h = 3.05 m, so the ball passes

$h' = 5.52- 3.05 = 2.47 m$

above the crossbar.

8 0

3 years ago

If you are at latitude 43 degrees north of Earth's equator, what is the angular distance (in degrees) from your zenith to the no

kirill115 [55]

Answer:

Your zenith is 43 N of 90 deg (equator)

Thus, your zenith is 90 - 43 = 47 deg

(At the N pole your zenith would be 0 deg from the N pole)

8 0

2 years ago

Phosphorus (P) is an element with an atomic number of 15 and an atomic mass of 31. How many neutrons are in an atom of phosphoru

BaLLatris [955]

(31-15 = 16).

Explanation:

the element phosphorus (P) has an atomic number of 15 and a mass number of 31. Therefore, an atom of phosphorus has 15 protons, 15 electrons, and 16 neutrons

5 0

3 years ago