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3 years ago

Two objects of different mass rest on Earth. Which one experiences the greater acceleration due to gravity ?

1 answer:

3 0

the answer is a because if two objects of diffrent mass are on tge earth, the object that would recive greather acceleration on gravity, wouldbe none.

it would be none cuz since two objects are on the earth, it ha sthe same gravitational piull, from the other object.

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A cylindrical metal specimen having an original diameter of 11.77 mm and gauge length of 46.1 mm is pulled in tension until frac

Marta_Voda [28]

Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area $= \frac{(A_1 - A_2)}{A_1 } * 100$

$A_1 =\pi r^2 = \pi * 5.88^2=108.8 mm2$

$A_2=\pi * 3.98^2= 49.97 mm2$

% reduction in area = $\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %$

b)percentage elongation $= \frac{(66 - 46.1)}{46.1} *100 = 43.16 %$

5 0

3 years ago

Which of the following objects is accelerating?

jenyasd209 [6]

Answer:

Explanation:

Constant speed (without change in direction) is not accelerating. If you are slowing down, speeding up, or changing direction, you are accelerating

4 0

3 years ago

What is cardiovascular eficiencia

Minchanka [31]

English: Cardiovascular efficiency depends on a number of factors. One measure is called stroke volume, which is the volume of blood pumped per heartbeat. A fit individual has a larger stroke volume, which means a greater volume of oxygen is delivered to the body per heartbeat.

Spanish: La eficacia cardiovascular depende de una serie de factores. Una medida se denomina volumen sistólico, que es el volumen de sangre bombeada por latidos cardíacos. Un individuo en forma tiene un volumen de movimiento mayor, lo que significa que un mayor volumen de oxígeno es entregado al cuerpo por latidos cardíacos.

5 0

4 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0

3 years ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

4 years ago

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