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2 years ago

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A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram p

rojectile is placed in the gun. The kinetic energy of the projectile when it is shot from the gun is

1 answer:

stepladder [879]2 years ago

4 0

Answer:

Explanation:

Given that,

Spring constant = 16N/m

Extension of spring

x = 8cm = 0.08m

Mass

m = 5g =5/1000 = 0.005 kg

The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.

So, Using conservation of energy

Energy in spring is converted to kinectic energy

So, Ux = K.E

Ux = ½ kx²

Then,

Ux = ½ × 16 × 0.08m²

Ux = 0.64 J

Since, K.E = Ux

K.E = 0.64 J

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Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at

BARSIC [14]

Answer: Option (c) is the correct answer.

Explanation:

Physical properties are the properties in which there is no change in chemical composition of a substance. On the other hand, chemical properties are the properties which change the chemical composition of a substance.

For example, when water boils at $100 ^{o}C$ then it changes into vapor state whereas when water freezes at $0^{0}C$ then it changes state from liquid to solid.

This means only physical state of water is changing and there is no change in chemical composition of water.

Hence, we can conclude that best option describing given information is that these are the physical changes water undergoes.

4 0

2 years ago

A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m Ã 3.00 m Ã 5.00 m. 1. what is the

cupoosta [38]

1) The volume of the raft is the product between the lenghts of its three dimensions:
$V = (2.00 m)(3.00m)(5.00m)=30 m^3$

2) The mass of the raft is the product between its density, d, and its volume, V:
$m=dV=(650 kg/m^3)(30 m^3)=19500 kg$

3) The weight of the raft is the product between its mass m and the gravitational acceleration, $g=9.81 m/s^2$ :
$W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N$

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
$W_a = W- \rho_W V_{disp} g$
where $\rho _W = 1000 kg/m^3$ is the water density and $V_{disp}$ is the volume of displaced fluid.
The density of the raft ( $650 kg/m^3$ ) is smaller than the water density ( $1000 kg/m^3$ ), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
$W=B=\rho _W V_{disp} g$
and therefore, the apparent weight will be zero:
$W_a = W-B=W-W=0$

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
$B=\rho_W V_{disp} g$
When the raft is completely immersed in the water, the volume of fluid displaced $V_{disp}$ is equal to the volume of the raft, $V_{disp}=V$ . Therefore the buoyancy in this situation is
$B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N$
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
$B=W=1.91 \cdot 10^5 N$

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
$mg = m_{disp} g$
where $m_{disp}= \rho_W V_{disp}$ is the mass of the displaced water. From the previous equation, we obtain that $m_{disp}=m=19500 kg$ .

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship $m=dV$ we can rewrite $m=m_{disp}$ as:
$d V =d_W V_{disp}$
and so
$V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3$

8) The volume of water displaced is (point 7) $19.5 m^3$ . This volume is now "filled" with part of the volume of the raft, therefore $19.5 m^3$ is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, $30 m^3$ :
$\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%$
Viceversa, the volume of raft above the water level is $30 m^3-19.5 m^3 = 10.5 m^3$ . Therefore, the fraction of volume of the raft above water level is
$\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%$

9) Let's repeat steps 5-8 replacing $\rho _W$ , the water density, with $\rho_E=806 kg/m^3$ , the ethanol density.

9-5) The buoyant force is given by:
$B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N$
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
$m_E = m = 19500 kg$

9-7) Using the relationship $d= \frac{m}{V}$ , we can find the volume of displaced ethanol:
$V_E = \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3$

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: $24.2 m^3$ . Therefore, the fraction of raft's volume below the ethanol level is
$\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%$
Consequently, the raft's volume above the ethanol level is
$30 m^3 - 24.2 m^3 = 5.8 m^3$
and the fraction of volume above the ethanol level is
$\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%$

8 0

2 years ago

For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0

2 years ago

In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg

harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 × $10^{-2}$ m

separation L = 0.5 mm = 0.5 × $10^{-3}$ m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 × $10^{-2}$ / 11

θ = 0.234

and we know diffraction grating for max

L sinθ = mλ

here we know m = 1 so put all value and find λ

L sinθ = mλ

0.5 × $10^{-3}$ sin(0.234) = 1 λ

λ = 2042.02 × $10^{-9}$ m

λ = 2042 nm

3 0

2 years ago

A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while

rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

$P_1V_1 =P_2 V_2$

Where,

$P_1$ and $V_1$ are the initial pressure and volume values

$P_2$ and $V_2$ are the final pressure volume values

The final pressure here is the atmosphere, then

$P_2 = 101325 \approx 1*10^5Pa$

$h = 10m$

$\rho_w = 1000kg/m^3$

$V_1 = 3.0L$

Pressure at the water is given by,

$P_1 = P_2 -\rho gh$

$P_1 = 1*10^5 +1000*9.8*10 =198000Pa$

Using Boyle equation we have,

$V_2 = \frac{P_1V_1}{P_2}$

$V_2 = \frac{198000*3*10^5}{10^5}$

$V_2 = 5.9L$

Therefore the volume of the lungs at the surface is 5.9L

4 0

1 year ago