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2 years ago

12

In a football game, a defensive lineman exerts a force upon an offensive lineman, causing the offensive lineman to be accelerate

d backward at 2 m/s2. If the offensive lineman's mass were doubled, how much force would the defensive lineman have to apply to accelerate the offensive lineman 2 m/s2 backward? A. half as much B. twice as much C. four times as much D. the same amount

2 answers:

Grace [21]2 years ago

7 0

the answer is B.twice as much

skelet666 [1.2K]2 years ago

7 0

Answer:

The answer is B (twice as much)

Explanation:

The magnitude, or amount, of an object's acceleration is directly proportional to the magnitude of the force exerted upon it and inversely proportional to its own mass.

In this case, if the offensive lineman's mass were doubled, then the force exerted upon him would also have to be doubled to produce the same acceleration (2 m/s2).

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viktelen [127]

^^^^^^^^^^^^^^^^^^^^^^^^^^^ is correct

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2 years ago

1. At a location in Europe, it is necessary to supply 1000 kW of 60-Hz power. Only power sources available operate at 50 Hz. It

Klio2033 [76]

Answer:

Explanation:

From the given information:

The speed of a synchronous motor in relation to its frequency can be represented with the formula:

$n_{sm}= \dfrac{120f_{se}}{P}$

where,

the electrical frequency $f_{se}$ is measured in Hz

the number of poles = P

For us to estimate the number of poles to have 50 Hz - 60 Hz Power, then we need to relate the frequencies of the above equation.

i.e

$\dfrac{120(50 \ Hz)}{P_1}= \dfrac{120( 60 \Hz)}{P_2} \\ \\ \dfrac{6000 \ Hz}{P_1}= \dfrac{7200 \ Hz}{P_2} \\ \\ \dfrac{P_2}{P_1}=\dfrac{7200}{6000} \\ \\ \\ \dfrac{P_2}{P_1}= \dfrac{12}{10}$

Thus, we can conclude that 10 poles synchronous motor is attached with 12 poles synchronous generator in order to convert 50 Hz to 60 Hz power.

3 0

2 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

2 years ago

An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o

jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substitute the values then we get

$-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}$

$\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}$

$\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}$

$v=-\frac{20}{3}=-6.7 cm$

Image distance of apple=-6.7 cm

Magnification=m= $\frac{v}{u}=\frac{-\frac{20}{3}}{-20}$

Magnification of apple= $\frac{1}{3}=0.33$

Hence, the magnification of apple=0.33

5 0

2 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

2 years ago