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crimeas [40]
2 years ago
14

In this vLab you used a complex machine to launch a projectile with the ultimate goal of hitting the target. Assume you built a

really big machine that could launch the projectile a "significant" distance; for instance, several hundred miles. Write a brief essay discussing the issues that would need to be accounted for with a projectile with that type of range. Be sure to include those issues affect the range of the projectile.
Physics
1 answer:
MissTica2 years ago
8 0
For a catapult to fire a projectile a significant range, the projectile will need a large mass. The machine would have to be very large to compensate for that. Also, the machine would be highly inaccurate. It would be entirely too difficult to pinpoint the exact location in which the projectile will hit. If you were to use a projectile that had a smaller mass, it would too easily be affected by friction, wind, and other outside forces. The machine used to fire the projectile itself, would have to be large, and it would be very inefficient.
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Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is
Scorpion4ik [409]

In a series circuit . . .

-- The total resistance is the sum of the individual resistors.

-- The current is the same at every point in the circuit.

The total resistance in this circuit is (3Ω + 6Ω )  =  9Ω

The current at every point is (V/R) = (12v / 9Ω ) = <em>1.33 A</em> .

Pick choice<em> (a)</em>.

6 0
1 year ago
13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?
sergeinik [125]

The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

Explanation:

The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of  mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

Q = mc del T

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

Q = 20*C*80

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

5 0
2 years ago
A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulu
Sergio [31]

Answer:

The gravitational acceleration on the planet is slightly less than g.

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

The formula can also be rewritten as

g=(\frac{2\pi}{T})^2 L (1)

In this problem, we have a pendulum which has a period of T=1.00 s on Earth. The length of the same pendulum must be shortened on the distant planet to have the same period of T'=1.00 s: this means that the length of the pendulum on the distant planet, L', is shorter than the length of the pendulum on Earth, L

L'

By looking at formula (1), we see that g (the gravitational acceleration) is directly proportional to L. therefore, if L is shortened on the distant planet, it means that also the value of g is lower than on Earth:

so, the correct answer is

The gravitational acceleration on the planet is slightly less than g.

7 0
2 years ago
Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0
2 years ago
When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

Force experienced, F=8\times 10^{-4}\ N

We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

3 0
2 years ago
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