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3 years ago

15

A 2 kg ball if at rest. If the ball accelerates to 20 m/s, what is the change in momentum?

1 answer:

katrin [286]3 years ago

4 0

Answer:

change in momentum

20m/s x2=40kg/m

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"The White Shark" allows riders to start from rest on a tube and then slide down a 44 meter slide. It takes the rider 6.2 second

Leni [432]

<span>Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculations of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:

d = v0t + at^2/2

We cancel the term with v0 since it is initially at rest,

d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2

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2 years ago

About how many centimeters will make an inch?<br> 02<br> O 10<br> 100<br> 200

Tresset [83]

There is approximately 2.54 cm that equals to 1 inch. So your closet answer would be the first choice. :)

7 0

3 years ago

Read 2 more answers

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa, т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P ) = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅ , where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅ of CD on the shaft BC

Torque = 660 * 0.3 cos∅

= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0

2 years ago

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Troyanec [42]

Answer:

The lowest possible frequency of sound is 971.4 Hz.

Explanation:

Given that,

Distance between loudspeakers = 2.00 m

Height = 5.50 m

Sound speed = 340 m/s

We need to calculate the distance

Using Pythagorean theorem

$AC^2=AB^2+BC^2$

$AC^2=2.00^2+5.50^2$

$AC=\sqrt{(2.00^2+5.50^2)}$

$AC=5.85\ m$

We need to calculate the path difference

Using formula of path difference

$\Delta x=AC-BC$

Put the value into the formula

$\Delta x=5.85-5.50$

$\Delta x=0.35\ m$

We need to calculate the lowest possible frequency of sound

Using formula of frequency

$f=\dfrac{nv}{\Delta x}$

Put the value into the formula

$f=\dfrac{1\times340}{0.35}$

$f=971.4\ Hz$

Hence, The lowest possible frequency of sound is 971.4 Hz.

8 0

2 years ago