Ask question

Ask question

All categories

3 years ago

7

A soccer ball was kicked. It had the mass of .42 kg and acceleration at 25m/s. what was the force?

2 answers:

IRINA_888 [86]3 years ago

6 0

Answer:

Force, F = 10.5 N

Explanation:

Given that,

Mass of the soccer ball, m = 0.42 kg

Acceleration of the soccer ball, $a=25\ m/s^2$

Let F is the force acting on the soccer ball. It can be calculated using second law of motion. Its mathematical form is given by :

$F=m\times a$

$F=0.42\times 25$

F = 10.5 N

So, the force acting on the soccer ball is 10.5 N. Hence, this is the required solution.

maw [93]3 years ago

4 0

Force of the kick or the balls velocity

You might be interested in

Select the correct answer.

aksik [14]

I don’t understand it I

4 0

4 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago

Why are overuse injuries particularly frustrating set-backs?

olganol [36]

Because they are caused by your exercise

7 0

3 years ago

Read 2 more answers

A tennis ball is hit at an angle of 20° to the horizontal with the velocity of 20 m/s. What is its velocity in the Y direction?

Alona [7]

B Hopefully I Helped... Good Luck

3 0

4 years ago

The surface area of a postage

ANTONII [103]

Answer:

608

Explanation:

Trust me

4 0

3 years ago

Read 2 more answers