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3 years ago

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some people agree that wearing seat belts reduces the risk of injurt to people in the car. however they say that the risk to oth

er road users might be increased. Suggest why it would be difficult to test the suggestion scientifically

1 answer:

leva [86]3 years ago

4 0

Answer:

It would be hard to test scientifically since it's subjective and can only be proven true if you conducted some experimentations and observations.

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A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the

const2013 [10]

Answer:

The equation for the object's displacement is $u(t)=0.583cos11.35t$

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

$k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in$

The angular speed is:

$w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s$

The damping force is:

$F_{D} =cu$

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

$c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in$

The critical damping is equal:

$c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in$

Like cc>c the system is undamped

The equilibrium expression is:

$u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t$

3 0

3 years ago

What are the principals of electric arc welding

Wewaii [24]

Principles<span> of </span>arc welding<span>. </span>Arc welding<span> is a </span>welding<span> process, in which heat is generated by an </span>electric arc<span> struck between an electrode and the work piece. </span>Electric arc<span> is luminous</span>electrical<span> discharge between two electrodes through ionized gas.</span>

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3 years ago

Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s

lana [24]

The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
</span><span>
Keisha follows the instructions for a demonstration on gas laws.
1. Place a small marshmallow in a large plastic syringe.
2. Cap the syringe tightly.
3. Pull the plunger back to double the volume of gas in the syringe.

Now, this activity is being done at the same temperature, because there is no mention of the temperature change. Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

7 0

4 years ago

Read 2 more answers

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. I

oksano4ka [1.4K]

Answer:

The deviation in path is $4.39 \times 10^{-3}$

Explanation:

Given:

Velocity $v = 1 \times 10^{6}$ $\frac{m}{s}$

Electric field $E = 500$ $\frac{V}{m}$

Distance $x = 1 \times 10^{-2}$ m

Mass of electron $m = 9.1 \times 10^{-31}$ kg

Charge of electron $q = 1.6 \times 10^{-19}$ C

Time taken to travel distance,

$t = \frac{x}{v}$

$t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }$

$t = 10^{-8}$ sec

Acceleration is given by,

$F = qE$

$ma = qE$

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }$

$a = 8.77 \times 10^{13}$ $\frac{m}{s^{2} }$

For finding the distance, we use kinematics equations.

$y = vt + \frac{1}{2} at^{2}$

Where $v = 0$ because here initial velocity zero

$y = \frac{1}{2} at^{2}$

$y = \frac{1}{2} \times 8.77 \times 10^{13 } \times (10^{-2} )^{2}$

$y = 4.39 \times 10^{-3}$ m

Therefore, the deviation in path is $4.39 \times 10^{-3}$

6 0

3 years ago

Which of the following actions in a car cause it to accelerate? Select all that apply.

Oksi-84 [34.3K]

Answer:

Option (b) will be correct answer for accelerating the car

Explanation:

For accelerating the car we have to speed up the car

In option (b) it is given that stepping on the gas pedal to speed up the car so it is correct answer

In option (c) it is given that pressing the pedal to slow down the car it is a case of deceleration so it is a incorrect statement.

In option (d) it is given that speed is constant at curved path so acceleration will be zero because if there is no change in velocity then acceleration will be zero

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3 years ago

Read 2 more answers