Can someone help me with a)

kirill115 [55]

–0.05 m/s

Explanation:

The total momentum of the system player+basketball must be conserved before and after the ball has been thrown.

Before throwing the ball, the total momentum of the system is zero, because can assume both the player and the basketball being at rest:

$p_i$

The total momentum after the ball has been thrown is instead the sum of the momenta of the the player and of the basketball:

$p_f=m_p v_p + m_b v_b$

where

$m_p = 59 kg$ is the player's mass

$v_p$ is the player's velocity

$m_b=0.51 kg$ is the ball's mass

$v_b=6 m/s$ is the ball's velocity

For the conservation of momentum, we have

$p_i=p_f$

$0=m_p v_p + m_b v_b$

$v_p=-\frac{m_b v_b}{m_p}=-\frac{(0.51 kg)(6 m/s)}{59 kg}=-0.05 m/s$

And the negative sign means that the player travels in the opposite direction to the ball.

8 0

3 years ago

Read 2 more answers

During spring tide, the sun, earth, and moon are in a straight line. This causes ...............

enyata [817]

Answer:

This causes higher average tidal ranges. The gravitational pull of the Sun and moon on Earth combined cause high tides that will be higher and low tides that will be lower than average.

4 0

3 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 81.7 cm and diameter 2.70 cm fr

Genrish500 [490]

Answer:

35.792 N

Explanation:

d = Diameter of rod = 2.7 cm

h = Length of rod = 81.7 cm

$\rho$ = Density of rod = 7800 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Volume of rod

$V=\dfrac{1}{4}\pi d^2h\\\Rightarrow V=\dfrac{1}{4}\times \pi\times (2.7\times 10^{-2})^2\times 81.7\times 10^{-2}$

Mass is given by

$m=\rho V\\\Rightarrow m=7800\times \dfrac{1}{4}\times \pi\times (2.7\times 10^{-2})^2\times 81.7\times 10^{-2}\\\Rightarrow m=3.6486\ kg$

Weight is given by

$W=mg\\\Rightarrow W=3.6486\times 9.81\\\Rightarrow W=35.792\ N$

The weight of the rod is 35.792 N

7 0

3 years ago

To keep the calculations fairly simple, but still reasonable, we shall model a human leg that is 92.0 cm long (measured from the

lorasvet [3.4K]

Answer:

Xc = 64.27 cm, Yc = 0 cm, Xc = 27.73 cm, Yc = 18.27 cm

Explanation:

A human leg is 92 cm long equally divided into two parts (upper leg and lower leg) that is both are 46 cm long

M1 = 8.35 kg

M2 = 5.50 kg

A. Starting from hip joint, when the leg is placed horizontally, the X coordinate of centre of mass will be

Xc = (m1x1 + m2x2) / m1 + m2

Xc = (8.35x46 + 5.50 x 92) / 8.35 + 5.50

( here we took x2 = 92 cm because we are measuring length from the hip joint, that means length at hip joint is 0 and at the foot 92 cm)

Xc = 890.1 / 13.85

Xc = 64.27 cm

B. Yc = (y1m1 + y2m2) / m1 + m2

Yc = (0x8.35 + 0x5.50) / 8.35 + 5.50

( here y1 and y2 are 0 because the leg is placed horizontally that means in X direction so none of the leg component is in Y direction)

Yc = 0 cm

C. Xc = (m1x1 + m2x2) / m1 + m2

Xc = (8.35x46 + 5.50x0) / 8.35 + 5.50

(here x2 is taken 0 because the other half leg is bent vertically that means it is in Y direction so its X component is 0)

Xc = 384.1 / 13.85

Xc = 27.73 cm

D. Yc = (m1y1 + m2y2) / m1 + m2

Yc = (8.35x0 + 5.50x46) / 8.35 + 5.50

(here y1 is zero because the upper leg is in X direction and y2 is taken 46 cm because starting measuring from hip joint 46 cm of length of lower leg is in Y direction)

Yc = 253/13.85

Yc = 18.27 cm

7 0

3 years ago

Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the

erik [133]

Answer:

Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).

Explanation:

5 0

3 years ago

Can someone help me with a)​

Can someone help me with a)