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4 years ago

Which of these observations would show a chemical

Physics

1 answer:

11Alexandr11 [23.1K]4 years ago

6 0

Answer: I'm pretty sure it'd be that lime juice burns your tongue

also where is choices a and b??

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A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc

alexdok [17]

Answer:

20.25 m

Explanation:

Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path.

That is;

ac = v²/r

Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m

Therefore;

r = v²/ac

= 27²/36

= 20.25 m

Hence the radius is 20.25 meters

5 0

3 years ago

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Evaluate the scenarios and select the one that demonstrates the training principle overload.

matrenka [14]

Answer:

B :)

Explanation:

:) JUST TRUST ME I GOT IT CORRECT

5 0

3 years ago

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The force between two magnetic poles varies directly with the distance of separation. True False

max2010maxim [7]

This is true because they are the same type of pole.

3 0

3 years ago

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A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kg
g = 9.8 m/s²
T = ?

Use formula:

$\boxed{\bold{T=m*g}}$

Replace and solve:

$\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}$

$\boxed{\boxed{\bold{T=98\ N}}}$

The tension in the rope is 98 Newtons.

Greetings.

5 0

3 years ago

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr

DiKsa [7]

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

$x_A = v_0 t + \frac{1}{2} at^2$

$v_0* t$ = velocity ( time ) = distance

since ball is at height, the above equation changes as

$x_A = H - \frac{1}{2} gt^2$

for ball B

$xB = v0 t - \frac{1}{2} gt^2$

the condition of collision is

xA = xB

$vA = - 2vB$ (given)

from the kinematic equation

the speed of the ball A is

$v_A = u- gt$

since initial speed of the ball A is zero

, so

$v_A = -gt$

the speed of the ball B is

$v_B = v_0 - gt$

since $v_A = - 2v_B$

$-gt = -2 ( v_0 - gt)$

$-gt = -2 v_0 +2gt$

$3gt =2 v_0$

$t = \frac{2v_0}{3g}$

since $x_A = x_B$

$H - \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2$

$H = v_0 t$

$= v_0 (2v_0/3g)$

$= \frac{2 v_0^2}{ 3g}$

$x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2$

$= 4 \frac{v_0^2}{9g} = (2/3) H$

so, (2/3) times height collision occur

3 0

4 years ago