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Alexxx [7]
3 years ago
9

What is the force if the mass is 75kg and the acceleration is 24.5m/s^2

Physics
1 answer:
Snowcat [4.5K]3 years ago
4 0

<u>Given;</u>

mass m = 75 kg

acceleration a = 24.5 ms²

<em>F = ma </em>

F  =  75 kg * 24.5 ms²

    =  1837.5 kg ms².

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The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;
Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0
2 years ago
Identify global climate zones and characteristics of each
IgorC [24]

research the different time zone around the world, and characteristics of each

3 0
2 years ago
Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete
harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

v=\dfrac{d}{T}

d is distance, d=2\pi r and 1/T = f (frequency)

v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s

So, the average speed of the rubber stopper is 18.84 m/s.  

4 0
3 years ago
Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi
insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

d

s−15

=

7

3

return trip is then :

d

s+15

=

7

5

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

8 0
3 years ago
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