The current shape of the solar system is *

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]

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.

dy = -1.5 m

a = 9.8m/s^2

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

t = ?

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

$dy = v_{iy}t + \frac{1}{2} at^{2}$

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

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

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

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

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

$0.553s = t$

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.

7 0

4 years ago

Eric has a mass of 80 kgkg. He is standing on a scale in an elevator that is accelerating downward at 1.7 m/s2m/s2. What is the

kolezko [41]

Answer:

The answer will be 936 N.

Explanation:

Given that

m = 80 kg

Acceleration of the elevator , a= 1.7 m/s² ( upward)

The gravity force on the mass = m g

The reading on the scale = F N

Now by applying the Newton's second law

F - m g = ma

F= m g + m a

F= m ( g +a )

F= 80 ( 10 + 1.7 ) N ( take g= 10 m/s²)

F=80 x 11.7 N

F= 936 N

Therefore the reading on the scale will be 936 N.

The answer will be 936 N.

5 0

3 years ago

A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?

Ilya [14]

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

= 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

8 0

2 years ago

If jack was traveling at 100 miles in 2 hours what was his velocity in miles per second

arsen [322]

1 hour is 3600 seconds. In 2 hours, it would be 7200 seconds. Divide the amount of miles by seconds. 100/7200=.01388..

6 0

3 years ago

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The velocity of a car is 65 m/s and it’s mass is 2515 kg. What is it’s KE?

Aloiza [94]

Example Problems. Kinetic Energy (KE = ½ m v2). 1) The velocity of a car is 65 m/s and its mass is 2515 kg. What is its KE? 2) If a 30 kg child were running at a rate of 9.9 m/s, what is his KE? Practice Problems. IN THIS ORDER…. Page 2: #s 6, 7, 8, 5. Potential Energy. An object can store energy as the result of its position.

4 0

3 years ago

Read 2 more answers