Ask question

Ask question

All categories

Murljashka [212]

2 years ago

7

A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball

1 answer:

DerKrebs [107]2 years ago

5 0

A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N

.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N

You might be interested in

Three 7 0hm resistors are connected in series across a 10 V battery. What is the equivalent resistance of the circuit?

Tomtit [17]

Given

Three 7 ohm resistor are in series.

The battery is V=10V

To find

The equivalent resistance

Explanation

When the resistance are in series then the resistance are added to find its equivalent.

Thus the equivalent resistance is:

$R=7+7+7=21\Omega$

Conclusion

The equivalent resistance is 21 ohm

4 0

8 months ago

You are falling off the edge.. What should you do to avoid falling..

Ugo [173]

If it is a matter of which way you are going you could lean forward. It would help to put all the weight opposite of where you are falling.

3 0

2 years ago

Read 2 more answers

A car is traveling south at a speed of 54 miles per hour and then begins traveling at a speed of 55 miles per hour but continues

Kazeer [188]

Explanation:

VELOCITY: BECAUSE ITS A VECTOR QUANTITY

7 0

2 years ago

Leonard is stressed and feeling overwhelmed by the college selection process. What should he do to combat this problem?

exis [7]

He should ask other people for advice

6 0

3 years ago

The diameter of a baseball is 7.4 cm and its mass is 0.15 kg. a) If a pitcher throws the baseball at a velocity of 44.3 m/s (100

Tema [17]

Answer:

drag force $F_D = 1.5 \ N$

Velocity (V) = 40.169 m/s

Explanation:

The drag force $F_D$ is given by the formula:

$F_D = C_D * \frac{1}{2}* \rho * V^2*A$

where:

$C_D$ = drag coefficient depending on the Reynolds number

Reynolds number Re = $\frac{\rho *V*D}{ \mu}$

Let's Assume that the air is in room temperature at 25 °C ; Then

density of the air $\rho$ = 1.1845 kg/m³

viscosity of fluid or air $\mu$ = 1.844 × 10⁻⁵ kg/ms

diameter of the baseball D = 7.4 cm

Velocity V = 44.3 m/s

Replacing them into the equation of Reynolds number ; we have :

$Re = \frac{1.1845 \ kg/m^3*44.3 m/s*0.074 m}{1.844*10^{-5}kg/ms}\\\\Re = 2.1*10^5$

A = Projected Area

From the diagram attached below which is gotten from NASA for baseball;

the drag coefficient which depends on Reynolds number is read as:

$C_D$ = 0.3

Projected Area A = $\frac{\pi D^2}{4}$

A = $\frac{\pi 0.074^2}{4}$

A = 0.0043 m²

Finally, drag force is then calculated as ;

$F_D = C_D * \frac{1}{2}* \rho*V^2*A\\\\F_D = 0.3* \frac{1}{2}*1.1845 \ kg/m^3*(44.3 \ m/s) ^2*0.0043 m^2\\\\F_D = 1.5 \ N$

b)

$- F_D = ma$

since acceleration a = $\frac{dV}{dt}$

Then;

$-F_D = m \frac{dV}{dt}$

Also;

velocity (V) = $\frac{dx}{dt}$

Then;

$- F_D = \frac{md_2x}{dt^2}$

$\frac{d_2x}{dt^2} = \frac {- F_D}{m}$

$F_D = 1.5 \ N\\m = 0.15 \ kg$

Then;

$\frac{d_2x}{dt^2} = \frac {- 1.5 }{0.15}$

$\frac{d_2x}{dt^2} =- 10$

Integrating the above equation ; we have :

$\frac{dx}{dt}= - 10 t + C\\$

when time (t) = 0 ; then $\frac{dx}{dt}= V = 44.3$

44.3 = - 10 × 0 + C

C = 44.3

$\frac{dx}{dt}= V = -10 t + 44.3$

Time (t) =

$\frac{distance }{velocity} \\\\= \frac{18.3 m}{44.3 m/s}\\\\= 0.413 s$

∴ Velocity ; $\frac{dx}{dt}= V = - 10t +44.3$

$\frac{dx}{dt}= V = - 10(0.413 s) +44.3$

Velocity (V) = 40.169 m/s

6 0

3 years ago