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andrew-mc [135]
2 years ago
12

Choose a system consisting only of ball A. What is the momentum change of the system during the collision?

Physics
1 answer:
shusha [124]2 years ago
3 0

The momentum change of the system during the collision is always equals to one another.

<h3>Momentum change during collision</h3>

In a collision, the momentum change of first object is equal to and opposite of the momentum change of second object with which the first one collided. Or in other words, the momentum lost by first object is equal to the momentum gained by second object.

The total momentum of the system means the collection of two objects is conserved so we can conclude that the momentum change of the system during the collision is always equals and opposite to one another.

Learn more about collisions here: brainly.com/question/7538238

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Is static electricity more harmful than it is helpful or vice versa
Deffense [45]

Answer:

You might even see a spark if the discharge of electrons is large enough. The good news is that static electricity can't seriously harm you. Your body is composed largely of water and water is an inefficient conductor of electricity, especially in amounts this small. Not that electricity can't hurt or kill you.

Explanation:

You might even see a spark if the discharge of electrons is large enough. The good news is that static electricity can't seriously harm you. Your body is composed largely of water and water is an inefficient conductor of electricity, especially in amounts this small. Not that electricity can't hurt or kill you.

6 0
3 years ago
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
3 years ago
A block weighs 15 n and is suspended from a spring that is attached to the ceiling. the spring stretches by 0.075 m from its uns
Illusion [34]

We can salve the problem by using the formula:

F=kx

where F is the force applied, k is the spring constant and x is the stretching of the spring.


From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:

k=\frac{F}{x}=\frac{15 N}{0.075 m}=200 N/m


By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied:

x=\frac{F}{k}=\frac{33 N}{200 N/m}=0.165 m

4 0
3 years ago
What force is needed to accelerate a 0.5kg football at a rate of 40m/s
GREYUIT [131]
Force=mass x acceleration
f= 0.5 x40
f=20N
6 0
3 years ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
3 years ago
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