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Alex777 [14]
3 years ago
15

Help me plz.Show workings​

Physics
2 answers:
Ksenya-84 [330]3 years ago
8 0

This question is a good example of the difference between 'speed' and 'velocity'.  This one is all 'velocity', and you've absolutely gotta keep the directions straight.

The object started out moving forward at 10 m/s.  It ended up moving backward at 2 m/s.  What was the change in its velocity ?

The change was <u>12 m/s backward</u> !  

Momentum = (mass) x (velocity).

Change in momentum = (mass) x (change in velocity)

This change in momentum = (0.1 kg) x (12 m/s backward)

This change in momentum = 1.2 m/s backward  or  <em>1.2 Ns  (B)</em>

ddd [48]3 years ago
3 0
Change in momentum: finial momentum - initial momentum
Momentum = mass * velocity
Mass = 100g, same as 0.1kg
m(v-u) = 0.1(10-2) = 0.1(8)
The answer is 0.8Ns
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Why do you think the temperature does not change much during a phase change?
11111nata11111 [884]

Answer:

It depends on where the temperature is dropping, in which body so to speak. Generally, the temperature adapts to the two bodies, for example if a hot piece of metal meets a cold one, the two will continue until they are at an equal temperature, an intermediate temperature.

3 0
2 years ago
Read 2 more answers
A 165 N object is supported by three cables(T1, T2 and T3), of which T1 and T2 are making angles θ1 = 52o and θ2 = 39o as shown
Liono4ka [1.6K]

Answer:

?Tension in string 2, T2 (in Newton) =

Answer for part 2

Hhgeppp plzzz

3 0
2 years ago
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

\frac{y-x}{y}=\frac{6}{15}

15 y - 15 x = 6 y

9 y = 15 x

y=\frac{5}{3}x

Differentiate with respect to time.

\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}

As given, dx/dt = 4 ft/s

\frac{dy}{dt}=\frac{5}{3}\times 4

\frac{dy}{dt}=\frac{20}{3} ft/s

6 0
3 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
Eduardwww [97]

Answer:

From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km

So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr

So runner towards the west will be

distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8

So equating east and west time we have

x/9= (11-x)/8

8x=99-9x

17x=99

x=5.92 km

That is the distance covered by runner towards the east and he will meet the runner toward the west at

6-5.92=0.08 km west of the flagpole.

7 0
3 years ago
An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 a
sveticcg [70]

Answer:

<h2>2.22 kPa</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{0.4 \times 500000}{0.9}   =  \frac{200000}{0.9} \\  = 222222.2222... \\  = 222222

We have the final answer as

<h3>2.22 kPa</h3>

Hope this helps you

5 0
3 years ago
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