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torisob [31]
2 years ago
11

Calculate the kinetic energy of a 2kg ball moving at 5m/s

Physics
1 answer:
shepuryov [24]2 years ago
8 0

Answer:

25

Explanation:

The kinetic energy is 25

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Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
Naddik [55]

Answer:

4.30 x 10⁵ N/C

Explanation:

Two positive +3.0-μC point charges at opposite corners of the square creates equal and opposite electric field at the center. hence the electric field by these two positive charges at opposite corners becomes zero.

a = length of the square of side = 0.50 m

r = distance of the center from each corner = \frac{a}{sqrt(2)} = \frac{0.50}{sqrt(2)} = 0.354 m

Magnitude of net electric field at the center is given as

E = \frac{2kq}{r^{2}} \\E = \frac{2(8.99\times10^{9})(3\times10^{-6})}{(0.354)^{2}} \\E = 4.30\times10^{5} NC^{-1}

6 0
3 years ago
a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. afte
IgorLugansk [536]
<span>From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block. 
m1 = 5.5g

5.5g x \frac{1kg}{1000g}= 0.0055kg

m2 = 22.6g

22.6g x \frac{1kg}{1000g}= 0.0226kg

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf. 

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x. 
</span>

dy = -1.5 m 

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

<span>We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity. 

dy = v_{iy}t +  \frac{1}{2} at^{2}</span>

1.5 = (0)t+\frac{1}{2} (9.8)t^{2}

<span>1.5 = \frac{1}{2} (9.8)t^{2}

\frac{(2)(1.5)}{9.8}=t^{2}

\frac{(3)}{9.8}=t^{2}

\sqrt{0.3061} = \sqrt{t^{2}

0.553s = t

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
v_{x} =d_{x}t and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
v_{x} =d_{x}t
v_{x} =(2.5m)(0.553s) 
v_{x} =4.52m/s

Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}

From this we can derive the equation for v1i by isolating it. 

v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}

Now let's put in all our given and what we solved:

v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}

v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}

v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}

v_{1i}= 23.09m/s

The initial speed of the dart is 23.09 m/s or 23.10 m/s.</span>
7 0
3 years ago
Which of these objects are malleable? Check all that apply.
g100num [7]

Answer:

The tin fork and knife, the copper coin, and the steel fence pole.

Explanation:

Those are both what people would call soft metals so they are malleable to the extent of probably not needing heavy duty equipment. It depends on you description of malleable because the steel fence pole could be malleable with the correct equipment and not snap in half if bent slowly enough.

The definition of malleable: (of a metal or other material) able to be hammered or pressed permanently out of shape without breaking or cracking.

But the glass table, marble sculpture and antique ceramic vase are nowhere near malleable because if you tried bending them they wouldn't bend but would shatter and break into pieces.

6 0
3 years ago
Read 2 more answers
You move a 25 n object 4 meters. find the work you did
d1i1m1o1n [39]
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:

Work = Force x distance
Work = 25 N x 4 meters
Work = 100 N.m</span>
5 0
3 years ago
Read 2 more answers
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
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