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andrew11 [14]
3 years ago
12

A table tennis ball with a mass of 0.003 kg and a soccer ball with a mass of 0.43 kg or both Serta name motion at 16 M/S calcula

te and compare the moments of both balls
Physics
2 answers:
poizon [28]3 years ago
8 0
Let us first know the given: Tennis ball has a mass of 0.003 kg, Soccer ball has a mass of 0.43 kg. Having the same velocity at 16 m/s. First the equation for momentum is P=MV P=Momentum M=Mass V=Velocity. Now let us have the solution for the momentum of tennis ball. Pt=0.003 x 16 m/s= (    kg-m/s ) I use the subscript "t" for tennis.  Momentum of Soccer ball Ps= 0.43 x 13m/s = (      km-m/s). If we going to compare the momentum of both balls, the heavier object will surely have a greater momentum because it has a larger mass, unless otherwise  the tennis ball with a lesser mass will have a greater velocity to be equal or greater than the momentum of a soccer ball.
Zepler [3.9K]3 years ago
7 0

Answer:

Momentum of tennis ball = 0.048 \frac{kg m}{s}

Momentum of soccer ball = 6.88 \frac{kg m}{s}

momentum of soccer ball is greater than that of tennis ball.

Explanation:

For table tennis ball,

Momentum of tennis the ball is given by,

Momentum = mass × volume

Momentum =  0.003 × 16

Momentum = 0.048 \frac{kg m}{s}

For table soccer ball,

Momentum of the soccer ball is given by,

Momentum = mass × volume

Momentum =  0.43 × 16

Momentum = 6.88 \frac{kg m}{s}

Thus, momentum of soccer ball is greater than that of tennis ball.

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UNO [17]

Answer: a) speed = 3.45 × 10^-2 m/s

b) speed = 1.38 × 10^-1 m/s

Explanation: shown in the attachment

3 0
3 years ago
2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or
krek1111 [17]

a) The work done by the tree is -1.44\cdot 10^5 J

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

W=Fd

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

W=-1.44\cdot 10^5 J is the work done

d=50.0 cm = 0.50 m is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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3 years ago
What radiation do remote controls use?
aleksandr82 [10.1K]

Answer:infrared radiation

Explanation:

Most remote control uses infrared radiation

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3 years ago
Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
3 years ago
A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p
Anettt [7]

Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

     u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

      u’ (1- u v / c²) = u -v

      u + u ’uv / c² = v - u’

      u (1 + u ’v / c²) = v - u’

      u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

      u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

      u = 0.51c / (1 + 0.4042)

      u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

       u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

       u = 1.29c / (1- 0.329)

       u = 0.961c

6 0
3 years ago
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