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3 years ago

To control whether an object is solid or incorporeal (things can pass through it) you would use the:

Physics

1 answer:

Zarrin [17]3 years ago

5 0

Answer:

Gamma radiation or Cathode rays

Explanation:

by striking incident gamma or cathode rays onto the solid when placed on a photographic plate

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A 600 g model rocket is on a cart that is rolling to the right at a speed of 4.0 m/s. The rocket engine, when it is fired, exert

murzikaleks [220]

Answer:

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

Explanation:

Step 1: Data given

mass of the rocket = 600 grams

speed = 4.0 m/s

Step 2: Calculate weight

Fw = mg

with Fw = the weight (in Newton)

with m = the mass (in kg)

with g = the acceleration due to gravity (9.81 m/s²).

Fw = (0.600 kg)(9.81 m/s²) = 5.886 N

Step 3: Calculate force available to provide acceleration

The rocket engine, when it is fired, exerts a 8.0 AND vertical thrust on the rocket.

5.886 N of that force will be used to counter the rocket's weight, leaving 2.114 N of force available to provide acceleration.

Step 4: Calculate the rocket's upward acceleration:

Fnet = m*a

With Fnet = the net force (the force that remains after the rocket's weight is compensated)

with a = the rocket's acceleration (in m/s²)

2.114 N = (0.600 kg)*a

a = 3.52 m/s² = the rocket's upward acceleration

Step 5: Calculate how long it will take to rise 20 meters into the air.

Δy = v0*t + 1/2 at²

with v0 = 0m/s

Δy = 1/2 at²

20 m = 1/2(3.52)t²

20 m = (1.76 m/s²)t²

11.36 = t²

t = 3.37 s

This means the rocket will take 3.37 seconds to reach the hoop. It should be launched when the cart is 3.37 seconds away from being directly beneath the hoop.

Step 6: Calculate the distance

v = Δx / t

4.0 m/s = Δx / 3.37 s

Δx = 13.48 m

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

8 0

4 years ago

A body of mass 1kg has a 25kgsmsqure moment of inertia about its center of mass. if it rotated about an axis which has a displac

ASHA 777 [7]

Answer:

194 kg m²

Explanation:

Use parallel axis theorem:

I = I₀ + mr²

I = 25 kg m² + (1 kg) ((5 m)² + (12 m)²)

I = 25 kg m² + 169 kg m²

I = 194 kg m²

6 0

3 years ago

A 4 kg block is pulled with 5.6 N of force to the right. The block experiences 1.1 N of friction. What is the acceleration of th

BARSIC [14]

Answer:

The acceleration of the box is 1.125 m/s² towards right.

Explanation:

Mass of the box, $m=4$ kg

Force acting towards right, $F=5.6$ N

Frictional force acting towards left, $f=1.1$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards right is given as:

$F_{net}=F-f=5.6-1.1=4.5\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\4.5=4a\\a=\frac{4.5}{4}=1.125\textrm{ }m/s^2$

Therefore, the acceleration of the box is 1.125 m/s² towards right.

7 0

4 years ago

For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red

barxatty [35]

Answer:

Explanation:

To calculate the red shift you use the following formula:

$z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1$

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

$z=\sqrt{\frac{c+v}{c-v}}-1$

- A red dwarf moving away from Earth at 39.1 km/s :

$z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}$

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

$z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}$

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

$z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}$

- A blue dwarf moving away from Earth at 25.9 km/s $z=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}$

- A red dwarf moving transversely at 14.1 km/s

$z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}$

6 0

4 years ago

Students of different ages were given the same puzzle to assemble. The puzzle assembly time was measured. The puzzle is meant fo

Crank

The older students should be able to complete the puzzle in a shorter amount of time than the younger students

6 0

3 years ago