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4 years ago

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Two equal forces act on two different objects, one of which has a mass ten times as large as the other. The larger object will h

ave _________ acceleration that the less massive object.

1 answer:

elena-s [515]4 years ago

4 0

Answer:

The larger object will have <u>smaller</u> acceleration that the less massive object.

Explanation:

Generally force is mathematically represented as

$F = ma$

=> $m = \frac{F}{a }$

at constant force we have

$m \ \alpha \ \frac{1}{a}$

So if m is increasing a will be decreasing which means the object with the larger mass will have less acceleration

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What do u do if ur baby sitting and the kid falls down the stairs and breaks his nose. Should I tell the parents or no? What do

igor_vitrenko [27]

Answer:

Call the parent's for the safety of the baby

Explanation:

5 0

4 years ago

Read 2 more answers

A bolt is dropped from a bridge under construction, falling 97 m to the valley below the bridge. (a) how much time does it take

exis [7]

The first thing we have to do for this case is write the kinematic equationsto
vf = a * t + vo
rf = a * (t ^ 2/2) + vo * t + ro
Then, for the bolt we have:
100% of your fall:
97 = g * (t ^ 2/2)
clearing t
t = root (2 * ((97) / (9.8)))
t = 4.449260429
89% of your fall:
0.89*97 = g * (t ^ 2/2)
clearing t
t = root (2 * ((0.89 * 97) / (9.8)))
t = 4.197423894
11% of your fall
t = 4.449260429-4.197423894
t = 0.252

To know the speed when the last 11% of your fall begins, you must first know how long it took you to get there:
86.33 = g * (t ^ 2/2)
Determining t:
t = root (2 * ((86.33) / (9.8))) = <span> 4.19742389 </span>s
Then, your speed will be:
vf = (9.8) * (4.19742389) = 41.135 m / s

Speed just before reaching the ground:
The time will be:
t = 0.252 + <span> 4.197423894</span> = <span> 4.449423894</span> s
The speed is
vf = (9.8) * (4.449423894) =<span> <span>43.603</span></span> m / s

answer
(a) t = 0.252 s
(b) 41,135 m / s
(c) 43.603 m / s

6 0

3 years ago

The length of the side of a cube having a density of 12.6 g/ml and a mass of 7.65 g is __________ cm.

qaws [65]

Density is the characteristic property of a substance. It is the measure of mass of the substance divided by its volume (density= mass/volume). Manipulate the given formula to come up with the formula for the volume. Therefore, volume is equals to mass of a substance divided by its density (Vol= mass/density). Given 12.6 g/ml as density and 7.65 g mass, volume is equals to 0.60714 ml, since 1 ml = 1cm^3, volume is equals to 0.60714 cm^3 then extract the cube root of the volume to get the length of the cube in cm which is equal to 0.84677 cm.

4 0

4 years ago

The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of

IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_{beat} = 0.99s$

Generally the frequency of the beat is

$f_{beat} = \frac{1}{t_{beat}}$

Substituting values

$f_{beat} = \frac{1}{0.99}$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_{beat}$ for $f_2$ having a higher tension

So

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_{beat}$

So

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}$

$T_2 = 0.99$ % lower than $T_1$

5 0

4 years ago

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car

marysya [2.9K]

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) = mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

4 0

3 years ago