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3 years ago

What is the acceleration of a 11 kg box sliding across the floor under the action of a net force 35N

Physics

1 answer:

Fofino [41]3 years ago

8 0

Answer:

<h3>The answer is 3.18 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

where

f is the force

m is the mass

We have

$a = \frac{35}{11} \\ = 3.181818...$

We have the final answer as

Hope this helps you

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A ball is thrown upward with a speed of 40 m/s. Approximately how much time does it take the ball to travel from the release loc

zvonat [6]

I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s². The solutions would be completely different if the same scenario were to play out in other places.

A ball is thrown upward with a speed of 40 m/s. Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.

Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip. After another 4.08 seconds, the ball has returned to the height of the hand which flung it. In total, the ball is in the air for <em>8.16 seconds</em> up and down.

4 0

3 years ago

A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height

stira [4]

Answer:

$9.5 kg m^2/s$

Explanation:

The angular momentum of an object is given by:

$L=mvr$

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

$L=(0.35)(9.0)(3.0)=9.5 kg m^2/s$

4 0

3 years ago

Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

olga_2 [115]

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

6 0

3 years ago

Although the Causes of Lunar Phases video is very useful for learning about phases of the Moon, it is inaccurate in some ways. W

lbvjy [14]

Answer:

In animations, usually the sizes of the objects are does not appear to be right because they are made to show the basic features and what's generally happening in the above mentioned case. These inaccuracies are consciously made for the better understanding of the audience. So following are the inaccuracies,

B.The astronaut is too big compared to the moon.

D.The moon is too big compared to its orbit.

F.The moon is too big compared to earth.

6 0

3 years ago

If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered

umka2103 [35]

Complete Question

The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ . The average rate of electric energy consumption in one house is 1.0 kW .

If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.

Answer:

The area is $A = 1.26 *10^{7} m^2$