What is evidence (and reasoning) on the Earth beginning as a small dense hot point (and beginning "13.7 billion years ago")

Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.

tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

4 0

2 years ago

In the two-slit experiment, monochromatic light of wavelength 600 nm passes through a 19) pair of slits separated by 2.20 x 10-5

kumpel [21]

Explanation:

It is given that,

Wavelength of monochromatic light, $\lambda=600\ nm=6\times 10^{-7}\ m$

Slits separation, $d=2.2\times 10^{-5}\ m$

(a) We need to find the angle corresponding to the first bright fringe. For bright fringe the equation is given as :

$d\ sin\theta=n\lambda$ , n = 1

$\theta=sin^{-1}(\dfrac{\lambda}{d})$

$\theta=sin^{-1}(\dfrac{6\times 10^{-7}}{2.2\times 10^{-5}})$

$\theta=1.56^{\circ}$

(b) We need to find the angle corresponding to the second dark fringe, n = 1

So, $d\ sin\theta=(n+\dfrac{1}{2})\lambda$

$sin\theta=\dfrac{3\lambda}{2d}$

$\theta=sin^{-1}(\dfrac{3\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{3\times 6\times 10^{-7}}{2\times 2.2\times 10^{-5}})$

$\theta=2.34^{\circ}$

Hence, this is the required solution.

4 0

3 years ago

An object oscillates back and forth on the end of a spring. Which of the following statements are true at some time during the c

Sindrei [870]

Answer:

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

Explanation:

For an object in simple harmonic motion, the total mechanical energy (sum of elastic potential energy and kinetic energy) is constant:

$E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$ (1)

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

We can also notice that the force on the spring is given by Hook's law:

$F=-kx$

And since according to Newton's law we have F = ma, this can be rewritten as

$ma=-kx\\a=-\frac{k}{m}x$

which means that the acceleration is proportional to the displacement.

So by looking again at eq.(1), we can now states that:

- when the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

- when the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

- in all the other intermediate situations, both velocity and acceleration are non-zero

So the correct answers are

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

8 0

2 years ago

A wooden bucket filled with water has a mass of 68 kg and is attached to a rope that is wound around a cylinder with a radius of

Aleonysh [2.5K]

Answer: 210.2N

Explanation:

Assume a bucket of water with a total mass of 68kg is attached to a rope, which in turn is tied around a 0.078m radius cylinder at the top of a well. A crank with a turning radius of 0.250 m is attached to the end of the cylinder.

the minimum force directed perpendicular to the crank handle required to raise the bucket is

(Assume the rope's mass is negligible, that the cylinder turns on friction-less bearings, and that g = 9.8 m/s2

The crank handle provides a torque T=0.25F where F is the force we are looking for.

A free body diagram will show that the tension in the rope times the cylinder radius R is equal to the torque on the cylinder. But the tension in the rope is just the weight of the bucket

W=mg= 68kg

W(0.078)=T=0.25F

F=0.312W=0.312(68kg)=21.216kg= 210.2N

7 0

3 years ago

Which of the following is an example of qualitative data

NemiM [27]

Qualitative data is easy to explain. You either agree or disagree in writing. Wrong. You agree, disagree, or qualify. Meaning agree to somewhat extend, but not entirely. So you can agree that people should walk instead of talk in the progressive manner, but not in the impulsive manner.

3 0

3 years ago