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viva [34]
3 years ago
15

Circle the statements that indicate a chemical property.

Physics
1 answer:
Lana71 [14]3 years ago
3 0
C) because a new substance is formed
B) not sure but might be because the chemical properties of the substance has changed
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How do you calculate acceleration
Eduardwww [97]

Answer: acceleration is equal to the change in velocity per unit time in seconds.

a= ∆v / t = vf - vi / t

Explanation: change in velocity or ∆v can be expressed as (vf - vi)

7 0
3 years ago
A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstr
Ksju [112]

Answer:

Explanation:

Given that

g=9.8m/s²

The spring constant is

k=50N/m

The length of the bungee cord is

Lo=32m

Height of bridge which one end of the bungee is tied is 91m

A steel ball of mass 92kg is attached to the other end of the bungee.

The potential energy(Us) of the steel ball before dropped from the bridge is given as

P.E= mgh

P.E= 92×9.8×91

P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

Uc= ½ke²

Where 'e' is the extension

Then the extension is final height extended by cord minus height of cord

e=hf - hi

e=hf - 32

Uc= ½×50×(hf-32)²

Uc=25(hf-32)²

Using conservation of energy,

Then,

The potential energy of free fall equals the potential energy in string

Uc=Us

25(hf-32)²=82045.6

(hf-32)² = 82045.6/25

(hf-32)²=3281.825

Take square root of both sides

√(hf-32)²=√(3281.825)

hf-32=57.29

hf=57.29+32

hf=89.29m

We neglect the negative sign of the root because the string cannot compressed

3 0
3 years ago
You and your little cousin sit on a see-saw. You sit 0.5 m from the fulcrum, and your cousin sits 1.5 m from the fulcrum. You we
svetoff [14.1K]

Answer:

200N

Explanation:

0.5/1.5=x/600N

1/3=x/600

x=200N

7 0
3 years ago
A pogo stick has a spring with a force constant of 2.50 × 104 N/m , which can be compressed 11.2 cm. what maximum height, in met
AleksandrR [38]

Answer:

h = 36.4 cm

Explanation:

given,

spring constant = 2.5 x 10⁴ N/m

compressed distance = 11.2 cm = 0.112 m

mass of the child = 44 kg

maximum height = ?

by energy of conservation

K.E_i +P.E_i= K.E_f + P.E_f

\dfrac{1}{2}kx^2 = mgh

\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h

\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h

156.8 = 44 \times 9.8 \times h

h = \dfrac{156.8}{44 \times 9.8}

h = 0.364 m

h = 36.4 cm

7 0
3 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
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