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2 years ago

A volleyball is hit and experiences a net

Physics

1 answer:

Ierofanga [76]2 years ago

6 0

Answer: 0.25 kg

Explanation:

Given that,

net force = 2 N

acceleration of volleyball = 8 m/s

mass of the volleyball = ?

According to the Newton 2nd law of motion, net force depends on the mass of the object and acceleration by which it moves.

i.e Net force = mass x acceleration

2N = mass x 8m/s²

Mass = 2N / 8m/s²

Mass = 0.25 kg

Thus, the mass of the volleyball is 0.25 kg

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You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it

VladimirAG [237]

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

4 0

2 years ago

You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$

height(h)=1 m

time taken=0.64 s

$h=ut+frac{at^2}{2}$

$1=0+\frac{a(0.64)^2}{2}$

$a=4.88 m/s^2and [tex]a=\alpha r$

where $\alpha$ is angular acceleration of pulley

$4.88=\alpha \times 5.2\times 10^{-2}$

$\alpha =93.84 rad/s^2$

And Tension in Rope

$T=m(g-a)$

$T=1.7\times (9.8-4.88)$

T=8.364 N

and Tension will provide Torque

$T\times r=I\cdot \alpha$

$8.364\times 5.2\times 10^{-2}=I\times 93.84$

$I=0.463\times 10^{-2} kg-m^2$

$I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2$

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0

2 years ago

A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

$v_x = \frac{x}{t}$