All categories

3 years ago

How did Naya simulate the force exerted on the head?

Physics

1 answer:

emmainna [20.7K]3 years ago

6 0

Answer:

Naya used a machine that shoots tennis balls, like this one, to fire balls at a mannequin's head. ... Once Naya knew how much the mannequin moved with each hit, she could calculate how much force the head had received from each blow.

You might be interested in

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X

Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

$u'=0.625 c$

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0

3 years ago

Mass m1 = 11.7 kg is on a horizontal surface. Mass m2 = 5.35 kg hangs freely on a rope which is attached to the first mass. The

Eddi Din [679]

Answer:

The magnitude of the system's acceleration is 2.362 m/s².

Explanation:

Given that,

Mass of one m₁= 11.7 kg

Mass of another m₂=5.35 kg

Coefficient of static friction = 0.574

Coefficient of kinetic friction = 0.106

We need to calculate the acceleration

Using balance equation

For m₂,

$m_{2}g-T=m_{2}a$

$T=m_{2}g-m_{2}a$ ...(I)

For m₁,

$T-f=m_{1}a$ ...(II)

Put the value of T in equation (II)

$m_{2}g-m_{2}a-\mu_{k}m_{1}g=m_{1}a$

$a=\dfrac{m_{2}g-\mu m_{1}g}{m_{1}+m_{2}}$

Put the value into the formula

$a=\dfrac{5.35\times9.8-0.106\times11.7\times9.8}{11.7+5.35}$

$a=2.362\ m/s^2$

Hence, The magnitude of the system's acceleration is 2.362 m/s².

8 0

4 years ago

Why does the pattern of colors repeat in a thin soap film? Please use 2 content related sentences

alexgriva [62]

Answer:

1) Some light waves bounce off the inner part of the film.

2) In doing so they travel a bit than the waves bouncing off the outer waves.

so we see colors repeat in pattern in a thin soap film

Explanation:

A colorful interference pattern is observed when light is reflected from the top and bottom boundaries of a thin oil film. The different bands form as the film's thickness diminishes from a central runoff-point.

When white light shines on a bubble, strange things happen. When light waves hit a bubble, some of them bounce straight back off the outer part of the soap film. Others carry on through but then bounce off the inner part of the film. So one set of light rays shine into a soap bubble, but two sets of rays come back out again. When they emerge, the waves that bounce off the inner film have traveled a tiny bit further than the waves that bounced off the outer film. So we have two sets of light waves that are now slightly out of step. Like two sets of ripples on a pond, these waves start merging. Just like on a pond, some add together and some cancel out. The overall effect is that some of the colors in the original white light disappear altogether, leaving other colors behind. These are the colors you see in soap bubbles.

5 0

4 years ago

A piece of copper wire is formed into a single circular loop of radius 9.1 cm. A magnetic field is oriented parallel to the norm

bazaltina [42]

Given Information:

Radius of circular loop = r = 9.1 cm = 0.091 m

Change in time = Δt = 0.66 seconds

Change in magnetic field = ΔB = 0.90 T

Resistance of wire per unit length = R = 2.9x10⁻² Ω/m

Number of turns = N = 1

Required Information:

Electrical energy dissipated = E = ?

Answer:

Electrical energy dissipated = 50.09x10⁻³ Joules

Step-by-step explanation:

We know that energy is given by

E = Pt

Where power is given by

P = ξ²/R

Where ξ is the induced EMF in the wire and is given by

ξ = -NΔΦ/Δt

Where ΔΦ is the change in flux and is given by

ΔΦ = ΔBAcosφ

Where φ is the angle between magnetic field and circular loop

A = πr² and R = 2.9x10⁻²*2πr

Substituting the above relations into the energy equation and simplifying yields,

E = [-Nπr²cosφ(ΔB/Δt)²]*t/R

E = [-1*π(0.091)²*cos(0)(0.90/0.66)²*0.66]/2.9x10⁻²*2π*(0.091)

E = 0.050094 Joules

E = 50.09x10⁻³ Joules

Therefore, the average electrical energy dissipated in the circular loop of the wire is 50.09x10⁻³ Joules.

6 0

3 years ago

A 5kg mass at rest on the ground is rised up to 15m. Find the work that was done on the object.

noname [10]

A. Before work is the answer

8 0

3 years ago

Read 2 more answers