a man counts 6 waves on a pond in 10 seconds. the distance between them is 40 cm. what is their speed?

A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a

vodka [1.7K]

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be

v_ox=v_o*cosФ

=60*cos (30)

= 51.96 m/s^-1

3 0

3 years ago

Read 2 more answers

If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

leonid [27]

Answer:

$\theta = tan^{-1}\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

$F_f = mgsin\theta$

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

$F_f = \mu N$

$N = mg cos\theta$

so we have

$F_f = \mu (mg cos\theta)$

so we will have

$mg sin\theta = \mu (mg cos\theta)$

so now we have

$tan\theta = \mu$

so maximum possible angle of the inclined plane is

$\theta = tan^{-1}\mu$

3 0

3 years ago

A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp

alex41 [277]

Answer:

5 m/s2, left

Explanation:

We can solve the problem by applying Newton's second law of motion, which states that:

$\sum F=ma$

where:

$\sum F$ is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have:

$\sum F=30 N - 20 N = 10 N$ (to the left) is the net force on the object

m = 2.0 kg is the mass

So, the acceleration is:

$a=\frac{\sum F}{m}=\frac{10}{2.0}=5.0 m/s^2$

in the same direction as the force (left).

5 0

3 years ago

In a redox reaction the substance that accepts electrons is said to be

netineya [11]

The question "<span>In a redox reaction the substance that accepts electrons is said to be?" is a bit vague. By definition, a "redox" or "reduction" reaction is one where classified by a gain of electrons. On the other hand, if it is a loss of electrons, then it is an oxidation reaction.</span>

8 0

4 years ago

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

#SPJ4

8 0

1 year ago

a man counts 6 waves on a pond in 10 seconds. the distance between them is 40 cm. what is their speed?​

a man counts 6 waves on a pond in 10 seconds. the distance between them is 40 cm. what is their speed?