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2 years ago

A golfer strikes a 0.059-kg golf ball with a force of 290 N. If the ball moves with a velocity of 69 m/s,

Physics

1 answer:

Ymorist [56]2 years ago

4 0

Answer:

0.014s

Explanation:

Given parameters:

mass of golf ball = 0.059kg

force applied = 290N

velocity = 69m/s

initial velocity = 0m/s

Unknown:

Time of contact = ?

Solution;

We know that momentum is the quantity of motion of body possess;

Momentum = mass x velocity

Momentum = 0.059 x 69 = 4.1kgm/s

Also; impulse is the effect of the force acting on a body;

impulse = force x time = momentum

So;

Force x time = momentum

Time = $\frac{momentum}{force }$ = $\frac{4.1}{290}$ = 0.014s

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An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight

EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

$v_{perpendicular} = v cos20$

$v_{parallel} = v sin20$

so here we have

$v_{parallel} = 16.2 sin20$

$v_{parallel} = 5.5 m/s$

so its velocity along the incline plane will be 5.5 m/s

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3 years ago

A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it

fomenos

Answer:68.15m/s

Explanation:

Given:

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

Formula:

v₁²=v₁²+2a (x)

Set up:

= $\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)$

<h2>Solution:</h2><h2>68.15m/s</h2>

6 0

3 years ago

Why do people hate on people and hate country people. are you country

Dmitry_Shevchenko [17]

Answer:

why do people hate on people and hate country people. are you country

because they only think about. themselves and they also think that they are always right .

it's my point of view

hope it is helpful to you

3 0

2 years ago

Read 2 more answers

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

Given:

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

What is the relationship between the atoms of elements,its isotopes and its ions

klasskru [66]

An element refers to a collection of atoms having the same number of protons and electrons (an atomic number). In each element there is a different atomic number due to a different amount of protons in the nucleus.

An isotope is a variation of an element that contains a different number of neutrons, therefore adding weight to the atom.

An ion is a charged atom, and its charge shows how many electrons it needs to gain or lose in order to become stable.

6 0

3 years ago