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2 years ago

physical changes occur when existing bonds break and new bonds form

1 answer:

6 0

<span>A physical change is any change that alters the form or appearance of a substance but does not change it into another substance
-Some of the physical properties may be altered, but the chemical composition stays the same
-Examples: Bending, Crushing, Cutting, Melting, Freezing, Boiling</span>

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A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c

Leya [2.2K]

Answer:

$T=83.37N$

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

$\sum F_y: T-mg=F_c$

Where $F_c$ is the centripetal force and is given by:

$F_c=ma_c=m\frac{v^2}{r}$

Replacing and solving for T:

$T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N$

8 0

3 years ago

The force of gravity on a car driving on the surface of a planet is called

nata0808 [166]

The answer should be A) Mass.

8 0

2 years ago

Read 2 more answers

If a runner has a speed of 8.66m/s and runs for 46.2s what distance is covered? tv = d

kati45 [8]

Answer:

$\text{Using the formula: }v=\frac{d}{t}\\\therefore vt=d\\\text{Plug and chug:}46.2(8.66)=400.092\text{ metres}$

6 0

2 years ago

A uniform electric field of magnitude 7.0 ✕ 104 N/C passes through the plane of a square sheet with sides 5.0 m long. Calculate

Vadim26 [7]

Answer:

$1.52*10^6 Nm^2/C$

Explanation:

Given that:

Electrical field E = $7.0 * 10^{-4}N/C$

square side l = 5.0 m

Area A = 5.0 * 5.0

= 25.0 m²

Angle ( θ ) between area vector and E = (90° - 60°)

= 30°

The flux $\phi_E$ can now be determined by using the expression

$\phi_E$ = $E*A*Cos \theta$

$\phi_E$ = $7.0 * 10^{-4}N/C$ $*25.0m*Cos 30^0$

$\phi_E$ = $1515544.457 Nm^2/C$

$\phi_E$ = $1.52*10^6 Nm^2/C$

5 0

2 years ago

Read 2 more answers

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0

3 years ago

﻿physical changes occur when existing bonds break and new bonds form

physical changes occur when existing bonds break and new bonds form