Ask question

Ask question

All categories

pickupchik [31]

2 years ago

8

A 77.8 kg basketball player is running in the positive direction at 8.1 m/s. She is met head-on by a 99.8 kg player traveling at

6.9 m/s toward her. If the 99.8 kg player is knocked backwards at 3.5 m/s, what is the resulting velocity of the 77.8 kg player?

1 answer:

lilavasa [31]2 years ago

6 0

Answer:

-5.24 m/s

** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **

Explanation:

Hi!

We can solve this problem considering each player as a point particle and taking into account the conservation of linear momentum.

Since the 99.8 kg player is moving towards the 77.8kg, the initial total momentum is:

m1*v1_i + m2*v2_i = (77.8kg)(8.1 m/s) - (99.8kg)(6.9 m/s)

** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **

The final total momentum is equal to:

m1*v1_f + m2*v2_f = (77.8 kg)v1_f + (99.8 kg)(3.5 m/s)

The conservation of momentu tell us that:

m1v1_i + m2v2_i = m1v1_f + m2v2_f

Therefore:

v1_f =v1_i + (m2/m1)*(v2_i-v2_f)

v1_f = 8.1 m/s + (99.8 / 77.8) * (-6.9 - 3.5 m/s)

<u>v1_f = -5.24 m/s</u>

You might be interested in

Develop expression for equation for uniform motion in a straight line

frozen [14]

According to cuneiform tablets in the ancient world, straight lines cannot cross, and no motion in the world is not relative. Btw...I KNOW!!! GOT MILK???

4 0

2 years ago

The gas pressure inside a container decreases when

avanturin [10]

Answer:

When the volume increases or when the temperature decreases

Explanation:

The ideal gas equation states that:

$pV= nRT$

where

p is the gas pressure

V is the volume

n is the number of moles of gas

R is the gas constant

T is the gas temperature

Assuming that we have a fixed amount of gas, so n is constant, we can rewrite the equation as

$\frac{pV}{T}=const.$

which means the following:

- Pressure is inversely proportional to the volume: this means that the pressure decreases when the volume increases

- Pressure is directly proportional to the temperature: this means that the pressure decreases when the temperature decreases

8 0

3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

2 years ago

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:

<u>Given:</u>

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

2 years ago

Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?

san4es73 [151]

Answer: A JUST DID IT ON USA TESTPREP

Explanation:

5 0

2 years ago