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erastovalidia [21]
3 years ago
8

The sensor in the torso of a crash test dummy records the magnitude znd direction of the net force acting on the dummy.If the du

mmy is thrown forward with a force of 130.0N while simultandously being hit from the side wigh a force of 4500.0N,what force will the sensor report
Physics
2 answers:
patriot [66]3 years ago
8 0

Answer : The force will be 4501.9

We can see that, two forces acting on the dummy in two different direction.

We know that, here two forces are given in perpendicular direction with each other.

We know the force is the vector addition law so, we will use the Pythagoras theorem for the resultant of the vectors

Now, the net force is

F_{net} = \sqrt{{F_{1}}^{2}+ {F_{2}}^{2}}

Two forces is given,

F_{1} = 130.0 N

F_{2} = 4500.0 N

Now, the net force is

F_{net} = \sqrt{(130)^{2}+ (4500)^{2}}N

F_{net} = \sqrt{20266900}N

F_{net} = 4501.9 N

Hence, the force will 4501.9 N

Whiterose2 years ago
0 0

The answer, if we use the tangent function and Pythagorean theorem, 45002 Newtons of force at 80.35 degrees

Whiterose
2 years ago
edit: 4502 newtons at 80.35 degrees
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Phoenix [80]
This instrument is called a spring scale.
5 0
3 years ago
Read 2 more answers
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0
2 years ago
The displacement of a 500 g mass, undergoing simple harmonic motion, is defined by the function :
Delicious77 [7]

The maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.

Simple Harmonic Motion

The given equation of the simple harmonic motion is

x=3.5 sin (\frac{\pi }{2t} + \frac{5\pi }{4} )

Data;

ω = π/2

k = 1.254N/m

Solving this

\frac{dx}{dt} = -3.5 X \frac{\pi }{2} cos (\frac{x\pi t}{2}+\frac{5\pi }{4}  )

Let's calculate the maximum velocity.

V_{m} =\frac{3.5\pi }{2}

This is only possible when cos θ = -1

The maximum kinetic energy is

K_m =\frac{1}{2} mv^2 = \frac{1}{2} X \frac{500}{1000} X \frac{7^2\pi ^2}^{4} ^2

w^2 = \frac{k}{m} \\k = w^2m\\k = \frac{\pi ^2}{4} X \frac{500}{1000} \\k =1.254 N/m

Using the value of spring constant, we can find the maximum potential energy.

P.E =\frac{1}{2} k x^2\\P.E =\frac{1}{2} X 1.234 X 3.5^2 \\P.E = 7.56 J

The maximum potential energy is 7.56J

The maximum mechanical energy is equal to the sum of maximum potential energy and the maximum kinetic energy.

ME = K.E + P.E

ME = 7.56J

From the calculations above, the maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

Learn more on simple harmonic motion here;

brainly.com/question/15556430

#SPJ1

8 0
2 years ago
Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t
fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, M_O=32

Molar mass of hydrogen, M_H=2

We know ideal gas law as:

PV=nRT

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵n=\frac{m}{M}

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

PV=\frac{m}{M}.RT

P=\frac{m}{V}.\frac{RT}{M}

        as: \frac{m}{V}=\rho\ (\rm density)

So,

P=\rho.\frac{RT}{M}

Now, according to given we have T,P,R same for both the gases.

P_O=P_H

\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}

\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}

\rho_O=16\rho_H

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0
3 years ago
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