What is the mass of an object that requires a force of 30 N to accelerate at a rate of 5 m/sec2 ?

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

3 years ago

Read 2 more answers

Anyone please help thank you

Thepotemich [5.8K]

Answer:

A) earth

B) live

C) live

D) earth

Explanation:

hope i help

4 0

2 years ago

a bowling ball of mass 7.3 kg and radius 9.6 cm rolls without slipping down a lane at 3.1 m/s . part a calculate its total kinet

erma4kov [3.2K]

49 J is the total kinetic energy. If a bowling ball of mass 7.3 kg and radius 9.6 cm rolls without slipping down a lane at 3.1 m/s. Kinetic energy is the energy an bowling ball has because of its motion.

Given: m = 7.3 Kg ; r = 9.4 cm = 0.094 m ; v = 3.1 m

Now total kinetic energy in this case is given by KE = Kinetic energy due to rotation + Kinetic energy due to translation

i,e KE = 1/2*m*v2 + 1/2*I*ω2 where I is the moment of inertia of the bowling ball about it's center and ω is the angular velocity

Now for pure rotation (without slipping) v = rω

also for the ball (solid sphere) I = 2/5*m*r2

Hence our kinetic energy becomes

KE = 1/2*m*v2 + 1/5*m*v2 = 7/10*m*v2

so KE = 0.7*7.3*(3.1)2 = 49.10 J = 49 J

Learn more about kinetic energy here

brainly.com/question/12669551

#SPJ4

5 0

1 year ago

A soccer player takes a cor- ner kick, lofting a stationary ball 35.0° above the horizon at 22.5 m/s. If the soccer ball has a m

liberstina [14]

Answer:

(a)

x=7.83 Kgm/s

y=5.48 Kgm/s

(b)

191.25 N

Explanation:

(a)

Change in momentum in x direction

$P_{x}=mvcos\theta$ where m is mass and v is the velocity and $\theta$ is the angle of kick

Substituting m=0.425 Kg, v=22.5m/s and $\theta=35^{o}$

$P_{x}=0.425*22.5*cos 35= 7.833141424$

$P_{x}=7.83 Kgm/s$

Change in momentum in y direction

$P_{y}=mvsin\theta$

$P_{y}=0.425*22.5*sin 35= 5.484824673$

$P_{y}=5.48 Kgm/s$

(b)

Force exerted by the player

F=mv/t where t is time

Substituting $t=5*10^{-2} s$

F=(0.425*22.5)/0.05= 191.25 N

8 0

2 years ago

When using a different calorimeter, and mixing 50 ml of hot water at 65 degrees c with 60 ml of water in the calorimete

Paha777 [63]

The specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

<h3>Conseervation of energy</h3>

The heat capacity of the calirometer is determined by applying the principle of conservation of energy.

Heat lost by the hot water = Heat gained by the calirometer

$Q _w = Q_c\\\\M_w C_w\Delta \theta _w = M_c C_c\Delta \theta _c$

where;

M is mass

mass = density x volume = ρV

Density of water = 1 g/ml

Mass of hot water = 1 x (50) = 50 g

Mass of water in calorimeter = 1 x (60) = 60 g

<h3>Equilibrium temperature</h3>

$\Delta T_c = 5.5\\\\T - 25 = 5.5\\\\T = 30.5 \ ^0C$

<h3>Specific heat capacity of the calirometer</h3>

$50 \times 4.184 \times (65 - 30.5) = 60 \times C_c \times (30.5 - 25)\\\\7217.4 = 330C_c\\\\C_c = \frac{7217.4}{330} \\\\C_c = 21.87 \ J/g^0C$

Thus, the specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

The complete question is below

When using a different calorimeter, and mixing 50 ml of hot water at

65 degrees C with 60 ml of water in the calorimeter at 25 degrees C, the temperature of the calorimeter increased by 5.5 degrees C.

a. Calculate the heat capacity of this calorimeter?

Learn more about heat capacity here: brainly.com/question/16559442

5 0

1 year ago