An average force of 328 N acts for a time interval of 0.05s on a golf ball

a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²

tigry1 [53]

At the player's maximum height, their velocity is 0. Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

which tells us the player's initial velocity $v_i$ is

$0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}$

The player's height at time $t$ is given by

$y=v_it-\dfrac g2t^2$

so we find their airtime to be

$0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s$

6 0

3 years ago

The largest hailstone every measured fell in Vivian, Nebraska in 2010. The circumference of that hailstone was 19 inches. Using

Crazy boy [7]

Answer:

6.04788 in

$115.82654\ in^3$

78.38779 m/s

0.88159 kg

34.55294 J

Explanation:

Circumference is given by

$c=2\pi r\\\Rightarrow r=\dfrac{c}{2\pi}\\\Rightarrow r=\dfrac{19}{2\pi}\\\Rightarrow r=3.02394\ in$

Diameter is given by

$d=2r\\\Rightarrow d=2\times 3.02394\\\Rightarrow d=6.04788\ in$

The diameter is 6.04788 in

$6.04788\times 2.54=15.3616152\ cm$

Volume of sphere is given by

$v=\dfrac{4}{3}\pi r^3\\\Rightarrow v=\dfrac{4}{3}\pi 3.02394^3\\\Rightarrow v=115.82654\ in^3$

The volume is $115.82654\ in^3$

$115.82654\times \dfrac{1}{1728}=0.06702\ ft^3$

Fall velocity is given by

$V=k\sqrt{d}\\\Rightarrow V=20\sqrt{15.3616152}\\\Rightarrow V=78.38779\ m/s$

The velocity of the fall will be 78.38779 m/s

Mass is given by

$m=\rho v\\\Rightarrow m=29\times 0.06702\\\Rightarrow m=1.94358\ lb$

$1.94358\ lb=1.94358\times \dfrac{1}{2.20462}=0.88159\ kg$

The mass is 0.88159 kg

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}0.88159\times 78.38779\\\Rightarrow K=34.55294\ J$

The kinetic energy is 34.55294 J

4 0

2 years ago

The longest banana split ever made was 7.32 km long (obviously they used more than one banana). If an archer were to shoot an ar

elixir [45]

Answer:

The horizontal displacement of the arrow is not larger than the banana split.

Explanation:

Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.

So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.

y - y₀ = ut - 1/2gt²

0 - y₀ = 0 × t - 1/2gt²

-y₀ = -1/2gt²

t² = 2y₀/g

t = √(2y₀/g)

Substituting the values of the variables, we have

t = √(2y₀/g)

= √(2 × 8848.04 m/9.8 m/s²)

= √(17696.08 m/9.8 m/s²)

= √(1805.72 s²)

= 42.5 s

The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s

So, d = vt

= 100 m/s × 42.5 s

= 4250 m

= 4.25 km

Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.

8 0

2 years ago

Thermal energy is added to a sample, its particles move ?

8_murik_8 [283]

Yes, if the temperature increases, than that means the particles are moving faster. Temperature is the measure of movement of particles in an object or substance.

By thermal energy, you mean adding heat correct....? I'm not very good at this sort of thing, but I gave you what I have..

4 0

2 years ago

Your friend thinks that the escape speed should be greater for more massive objects than for less massive objects. Provide an ar

Brrunno [24]

Answer:

Concepts and Principles

1- Kinetic Energy: The kinetic energy of an object is:

K=1/2*m*v^2 (1)

where m is the object's mass and v is its speed relative to the chosen coordinate system.

2- Gravitational potential energy of a system consisting of Earth and any object is:

U_g = -Gm_E*m_o/r*E-o (2)

where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.

Solution

The argument:

My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.

The counterargument:

We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):

K=1/2*m*v^2 (1)

and a gravitational potential energy Ug given by Equation (2):

Ug = -G*Mm/R

where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:

K+U=0

1/2*m*v^2 + (-G*Mm/R) = 0

1/2*m*v^2 = G*Mm/R

1/2*v^2 = G*M/R

solving for v we get

v = √2G*M/R

so we see v does not depend on the mass of the projectile

8 0

3 years ago