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3 years ago

A 76.O kg person is being pufed away from a burning building as shown below

Physics

1 answer:

aleksandrvk [35]3 years ago

4 0

Answer:

Final Answer

T1=736 NT1=736 N

T2=194

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Which is an example of a physical change? wood rots. gasoline ignites. water evaporates. a nail rusts. description?

tester [92]

Wood Rots is the correct answer, as the wood begins to die

6 0

3 years ago

Read 2 more answers

What is the average speed of a train that travels at 100km/hr for 6 hours and then 120km/hr for 5 hours

Sindrei [870]

Answer:

Explanation:

Average speed = Total distance / Total time.

100 km/hr

r = 100 km / hr

t = 6 hours

d = 6 * 100 = 600 km

120 km / hr

r = 120 km / hr

t = 5 hour

d = 120 * 5

d = 600 km

Total distance = 600 + 600 = 1200 km

Total time = 5 hour + 6 hours = 11 hours.

Average speed = 1200 km / 11 hours = 109.1

5 0

3 years ago

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

3 years ago

As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate

Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current $I$ . Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field $B$ . The direction of the right thumb should now point in the direction of the force on the wire $F$ .)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

7 0

3 years ago

a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$

So,

Time in which train will come to rest = 20 seconds

4 0

3 years ago