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m_a_m_a [10]
1 year ago
9

A 1,200 kg car accelerates at 3 m/s2. What net force is the car experiencing during thisacceleration?0 3600 NO 2000 N0 2400 NOON

Physics
1 answer:
Mama L [17]1 year ago
5 0

m = mass = 1,200 kg

A = acceleration = 3 m/s^2

Apply Newton's second law:

Force = mass x acceleration

F = 1,200 x 3 =3600 N

The net force the car experiences is 3600 N

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In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is
meriva

The spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

<h3>What is double slit experiment?</h3>

The double-slit experiment serves as a proof in current physics that both light and matter may exhibit properties of classically defined waves and particles. It also illustrates the inherently probabilistic nature of quantum mechanical events. Thomas Young carried out the first experiment of this kind employing light in 1802, illustrating how light behaves like a wave. It was formerly believed that light was made up of either waves or particles. About a century later, with the advent of modern physics, it was discovered that light may in fact exhibit behaviour like that of both waves and particles. The identical behaviour of electrons was first shown by Davisson and Germer in 1927, and it was later extended to atoms and molecules.

The separation between the slits, d = 0.05mm = 5×10⁻⁵ m

The distance from the slits to a screen, D = 2.5 m

Let x is the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits,

λ = 600nm = 6× 10⁻⁷ m  

We know that the bright fringe is given by :

y = nλD/d

So, the spacing between the first-order and second-order bright fringes is :

x = 2λD/d - λD/d

x =  λD/d

x = 6 × 10⁻⁷ × 2.5/5 × 10⁻⁵

x = 0.03 m

or

x = 3 cm

So, the spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

to learn more about double slit experiment go to -

brainly.com/question/28108126

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8 0
1 year ago
Chris threw a basketball a distance of 27.5 m to score and win his
salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}

<u>v₀ = 16.55 m/s</u>

8 0
3 years ago
Lauren rode the bus home from the mall on Saturday. She felt the bus accelerate as it left the parking lot, and then the bus tur
Alex_Xolod [135]
A. is the one it did not say
7 0
3 years ago
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
The eccentricity of an asteroid's orbit is 0.0442, and the semimajor axis is 1.12 x 1011 m. The Sun's center is at one focus of
nadezda [96]

Answer:

a. d=99x10^{8}m

b. r=14.22 R_s

Explanation:

The eccentricity of an asteroid's is 0.0442 so

a.

to find the focus distance between both focus is

d=2*e*a

e=1.12x10^{11}m\\a=0.0442

So replacing numeric

d=2*1.12x10^{11}m*0.0442=9900800000

d=99x10^{8}m

b.

Now to find the ratio of that distance between the solar radius and the distance

r=\frac{d_1}{d_s}

r=R_s*\frac{99x10^8m}{6.96x10^8m}

r=14.22 R_s

5 0
4 years ago
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