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3 years ago

A wagon has a mass of 100 grams and an

Physics

1 answer:

Elza [17]3 years ago

7 0

Answer: 0.2N

Explanation:

mass of wagon = 100 grams

Convert mass in grams to kilograms

If 1000 grams = 1kg

100 grams = (100/1000) = 0.1kg

Acceleration of wagon = 2m/s/s

Force on the wagon = ?

Recall that Force is the product of mass of an object by the acceleration by which it moves.

i.e Force = mass x acceleration

Force = 0.1kg x 2m/s/s

Force = 0.2N

Thus, the force on the wagon is 0.2Newton

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hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .

fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)

do =original position = 2 m

vo = original VERTICAL velocity = 0

a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t - 1/2 ( 9.81)t^2

to show t = .639 seconds to hit the ground

During this .639 seconds it flies horizontally at 10 m/s for a distance of

10 m/s * .639 s = 6.39 m

5 0

1 year ago

An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan

MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

$Vertical \ velocity \ (\frac{m}{s} ) = u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) = u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) = a_{v} = 9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) = a_{h} = 0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}$

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

$u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...$

7 0

3 years ago

Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (

lesya [120]

Answer:

a) When moving towards a high pressure center the pressure values increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.

When moving towards a high pressure center the pressure values increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

6 0

3 years ago

An object starts at rest. Its acceleration over 30 seconds is shown in the graph below:

ddd [48]

Answer:

The instantaneous speed of the object after the first five seconds is 12.5 m/s.

(C) is correct option.

Explanation:

Given that,

An object starts at rest. Its acceleration over 30 seconds.

We need to calculate the instantaneous speed of the object after the first five seconds

We know that,

Area under the acceleration -time graph gives speed.

According to figure,

$speed = area\ of\ tringle$

$speed=\dfrac{1}{2}\times b\times h$

$speed =\dfrac{1}{2}\times5\times5$

$speed-12.5\ m/s$

Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.

6 0

3 years ago

To cross the river a swimmer chooses the direction minimizing the amount of time spent in the water. The swimmer swims at a cons

Lynna [10]

Answer:

304 meters downstream

Explanation:

The given parameters are;

The speed of the swimmer = 2.00 m/s

The width of the river = 73.0 m

The speed of the river = 8.00 m/s

Therefore;

The direction of the swimmer's resultant velocity = tan⁻¹(8/2) ≈ 75.96° downstream

The distance downstream the swimmer will reach the opposite shore = 4 × 73 = 304 m downstream

The distance downstream the swimmer will reach the opposite shore = 304 m downstream

6 0

3 years ago