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kaheart [24]
2 years ago
8

7. Four perpendicular forces, 2 N, 3 N, 9 N and 10 N act on object P as shown in Diagram 2. 10 N A B 9 N 3 N 2 N P Diagram 2 Wha

t is the magnitude of the resultant force? 10 N C 18 N 14 N D 20 N​

Physics
1 answer:
Anvisha [2.4K]2 years ago
6 0

Answer: R=

A

2

+B

2

+2ABcosθ

​

θ=90

o

cos90

o

=0

R=

A

2

+B

2

​

=

3N

2

+4N

2

​

=5N

Explanation:

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A little girl slid down a playground slide, decreasing her potential energy by 1000 J while increasing her kinetic energy by onl
koban [17]

Answer:

Explanation:

This is because The 100 J of potential energy that doesn't go into increasing her kinetic energy goes into thermal energy—heating her bottom and the slide.

5 0
3 years ago
Read 2 more answers
Which type of graph would be best for showing the percentage of people in a family with different jobs? A. Circle graph B. Scatt
musickatia [10]

Answer:

A

Explanation:

A circle graph typically represents numbers in percentages, used to visualize a part to whole relationship or a composition

6 0
3 years ago
Do geochemists need to have a knowledge of physical science? Explain your<br> answer. please help:))
saul85 [17]
No they don’t lol I hate school
6 0
3 years ago
A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet
Leno4ka [110]

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

8 0
3 years ago
Fighter jet starting from airbase A flies 300 km east , then 350 km at 30° west of north and then 150km north to arrive finally
Tems11 [23]
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North. 
4 0
3 years ago
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