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4 years ago

12

What type of force does a backpack exert on the body’s back and hips?

1 answer:

Inessa [10]4 years ago

7 0

The type of force that <span>a backpack exerts on the body’s back and hips can be considered as a vertical force due to the force of gravity along with the load of things present in the backpack. The force is exerted vertically downwards. I hope that this is the answer that has actually come to your desired help.</span>

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A psychopathologist records the number of criminal offenses among teenage drug users in a nationwide sample of 1,201 participant

lara [203]

Answer:

Standard deviation = 3

Explanation:

Given

$N = 1201$

$SS = 10800$

Required

Determine the standard deviation

First, we need to determine the variance;

$Variance = \frac{SS}{N - 1}$

This gives:

$Variance = \frac{10800}{1201 - 1}$

$Variance = \frac{10800}{1200}$

$Variance = 9$

Know that:

$Variance = SD^2$

Where SD represents standard deviation

This gives

$9 = SD^2$

Take square root

$SD = \sqrt 9$

$SD = 3$

7 0

3 years ago

Which of the following statements is accurate? A. Sound waves passing through the air will do so as transverse waves, which vibr

egoroff_w [7]

The answer is, C. the wavelength is measured in parallel to the direction of the wave, at any point, under the same repetition for any type of wave.

7 0

4 years ago

Read 2 more answers

Resistance is a force that __________.

melisa1 [442]

The answer is slows objects down.

3 0

3 years ago

Read 2 more answers

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

3 years ago

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe

JulsSmile [24]

Answer:

(A) $a=2.0.37m/sec^2$

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $88km/hr=88\times \frac{1000}{3600}=24.444m/sec$

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So $24.444=0+a\times 12$

$a=2.0.37m/sec^2$

So acceleration of the car will be $a=2.0.37m/sec^2$

(b) From third equation of motion $v^2=u^2+2as$

So $24.444^2=0^2+2\times 2.037\times s$

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

6 0

4 years ago