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Crank
4 years ago
7

Help me ASAP (this is science ;-;)

Physics
2 answers:
IgorC [24]4 years ago
6 0
Go on .........I don’t see a question
krek1111 [17]4 years ago
5 0
There is no question on here
You might be interested in
A newly discovered particle, the SPARTYON, has a mass 945 times that of an electron. If a SPARTYON at rest absorbs an anti-SPART
Gennadij [26K]

Answer:

\nu =1166\times 10^{20}Hz  

Explanation:

We have given the rest mass of SPARTYON = 945 times of mass of electron

We know that mass of electron =9.11\times 10^{-31}kg

So mass of SPARTYON =945\times 9.11\times 10^{-31}=8608.95\times 10^{-31}kg

Speed of light c=3\times 10^8m/sec

According to Einstein equation energy is given by

E=mc^2=8608.95\times 10^{-31}\times (3\times 10^{8})^2=77480.55\times 10^{-15}j

Now according to planks's rule

Energy is given by

E=h\nu , here h is plank's constant h=6.6\times 10^{-34}

So 77480.55\times 10^{-15}=6.6\times 10^{-34}\nu

\nu =1166\times 10^{20}Hz  

3 0
3 years ago
Find the tension in the two ropes that are holding the 4.2 kg object in place.
Blababa [14]

Answer:

The tension in the two ropes are;

T1 = 23.37N T2 = 35.47N

Explanation:

Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity

W = 4.2×10 = 42N

The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.

Using triangular law of force and sine rule to get the tension we have;

If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;

T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}

T1/sin33.8° = T2/sin57.6° = 42/sin88.6°

From the equality;

T1/sin33.8° = 42/sin88.6°

T1 = sin33.8°×42/sin88.6°

T1 = 23.37N

To get T2,

T2/sin57.6°= 42/sin88.6°

T2 = sin57.6°×42/sin88.6°

T2 = 35.47N

Note: Check attachment for diagram.

7 0
3 years ago
Which optical instrument, a telescope, a microscope, or a camera, forms images in a way most like your eye? Explain.
Xelga [282]

Camera is an optical instrument that forms images in a way most like human eyes.

Explanation:

The eye consists of a small spherical globe of about 2 cm in diameter, which is free to rotate under the control of 6 extrinsic muscles. Light enters the eye through the transparent cornea, passes through the aqueous humor, the lens, and the vitreous humor, where it finally forms an image on the retina.

In case of a digital camera, CCD is used known as charged coupled device. CCD is actually in the shape of array or a rectangular grid. It is like a matrix with each cell in the matrix contains a censor that senses the intensity of photon. When light falls on the object , the light reflects back after striking the object and allowed to enter inside the camera which is then stored by CCD in the form of electric signals.

Similarities between eye and a camera:

Cornea and Lens - The cornea is the “cap” of the eye. This transparent (like clear jelly) structure sits to the front of the eye. The lens of a camera is also transparent (glass) and sits at the front of the body.  Both allows light to enter and have spherical curvature.

Iris and aperture -The aperture is to the camera as the iris is to the eye. Both control the amount of light allowed to enter the eye or camera.

Retina and Film - The retina sits at the back of the eye and collects the light reflected from the surrounding environment to form the image. The same task in the camera is performed either by film or sensors in digital cameras.

Keywords: eye, camera, CCD, iris, film, aperture, cornea, lens, image formation

Learn more about camera from brainly.com/question/1771913

#learnwithBrainly

3 0
4 years ago
A spring with k = 15.3 N/cm is initially stretched 1.81 cm from its equilibrium length. a) How much more energy is needed to fur
leonid [27]

Answer:

2.31J

Explanation:

the energy for a spring system is given by:

E=\frac{1}{2} kx^2

where k is the spring constant: k=15.3N/cm=1530N/m and x is the distance stretched from the equilibrium position.

In the first case x=1.81cm=0.0181m

so the energy to stretch the spring 1.81cm is:

K_{1}=\frac{1}{2} (1530N/m)(0.0181m)^2=0.25J

and for the second case,  the energy to stretch the spring 5.79cm:

x=5.79cm=0.0579m

K_{1}=\frac{1}{2} (1530N/m)(0.0579m)^2=2.56J

so to answer a) we must find the difference between these energies:

2.56J-0.25J=2.31J

6 0
4 years ago
what is the pressure on a circular helipad of diameter 100 m on which a helicopter of weight 40,000 N is landing. *
Luba_88 [7]

Answer:

5.09 Pa

Explanation:

Area of the circle = πr²

The diameter is given, so 100/2 = 50 m is the radius.

π x 50² = 7854 m²

Pressure = force/area

Pressure = 40000/7854

Pressure = 5.09 Pa

Hope this helps!

4 0
3 years ago
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