Describe how you determine whether an object is in motion

A single slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum, when monochromatic

Margarita [4]

Answer:

$\lambda = 424.4 nm$

Explanation:

As we know by the formula of diffraction

$a sin\theta = N\lambda$

so we have

a = slit size

$\theta$ = angular position of Nth minimum

so we will have

for first minimum of 630 nm light

$a sin40 = 1(630 \times 10^{-9})$

$a = 9.8 \times 10^{-7} m$

Now for another wavelength second minimum is at 60 degree angle

$a sin60 = 2 \lambda$

$(9.8 \times 10^{-7}) sin60 = 2 \lambda$

$\lambda = 424.4 nm$

3 0

3 years ago

Why aren't iron, cobalt and iron in the same group of elements?

aev [14]

Because iron is a metal and cobalt is a non-metal

3 0

3 years ago

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/

polet [3.4K]

V o - initial velocity
v = velocity at the maximum height,
v² = v o² - 2 g h
v = 0
0 = v o² - 2 g h
v o² = 2 g h = 2 · 9.80 · 0.460
v o² = 9.052
v o = √9.052 = 3.004197 m/s ≈ 3 m/s

8 0

3 years ago

Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu

morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

$U=mgh$

$U=2.35\ kg\times 9.8\ m/s^2\times 1.3\ m$

U = 29.93 Joules

So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

6 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$