Ask question

Ask question

All categories

3 years ago

12

1. A ball is launched horizontally at 40 m/s while 6.0 m above a level field . How far will the ball move horizontally before st

riking the ground ?

1 answer:

Tresset [83]3 years ago

5 0

Answer:

44 m

Explanation:

Given that,

Horizontal velocity of the ball, u = 40 m/s

It is 6 m above the level field.

We need to find the distance covered by the ball when move horizontally before striking the ground. Let it is d.

Firstly, we will find time taken for the ball to hit the ground. Using second equation of motion as follows :

$s=ut+\dfrac{1}{2}at^2$

Put u = 0 and a = g

$s=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2s}{g}} \\\\t=\sqrt{\dfrac{2\times 6}{9.8}} \\\\t=1.1\ s$

No finding the horizontal distance as follows :

d = ut

d = 40 m/s × 1.1 s

d = 44 m

So, the ball will move 44 m horizontally before striking the ground.

You might be interested in

Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy

erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14 + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

= 1.35 - .157

= 1.193 MeV.

5 0

3 years ago

Tanya [424]

Answer:

Reset

Explanation:

Digital methods are the methods that are uses methodological outlook to study societal change and cultural condition of online data. Reset is use to disguise data In digital methods. It is use to set again and conceal data by giving the data a different form. It restores the device to the original manufacture's settings.

8 0

3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. C

Vlada [557]

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

$W=-P\Delta V=-P(V_2-V_1)$

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm (760torr =1atm)

$V_1$ = initial volume = 5.68 L

$V_2$ = final volume = 2.35 L

Putting values in above equation, we get:

$W=-0.96atm\times (2.35-5.68)L=3.20L.atm$

To convert this into joules, we use the conversion factor:

$1L.atm=101.33J$

So, $3.20L.atm=3.20\times 101.3=324J$

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

5 0

3 years ago

45. A particle moves in a straight line with an initial velocity of 30 m/s and constant acceleration 30 m/s2 . (a) What is its d

omeli [17]

Answer:

Displacement after 5 seconds is 155/2 meters

Explanation:

Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.

Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.

Displacement in 5 seconds is given by X (5) - X (0).

X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2

Displacement after 5 seconds is 155/2 meters

7 0

3 years ago

Read 2 more answers