PLEASE HELP! I'LL GIVE BRAINLEST

A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th

enyata [817]

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

Mass of the child, m = 16 kg
Length of the incline, L = 2 m
Angle of inclination, θ = 20⁰

The vertical height of fall of the child from the top of the incline is calculated as;

$sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m$

The gravitational potential energy of the child at the top of the incline is calculated as;

$P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J$

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0

2 years ago

calculate the work done by 2N force directed at 30 degree to the vertical to move a 500g box a horizontal distance of 400 cm acr

sattari [20]

Answer:

there is no picture :o?

Explanation:

:oo

3 0

2 years ago

The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

Finally:

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

2 years ago

A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j

elena55 [62]

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

6 0

3 years ago

A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?

docker41 [41]

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

$v=\sqrt{2gh}$

g is acceleration due to gravity

$v=\sqrt{2\times 9.81\ m/s^2\times 2\ m}$

v = 6.26 m/s

So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.

8 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST​

PLEASE HELP! I'LL GIVE BRAINLEST