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amm1812
3 years ago
15

This company is run by kids right

Physics
2 answers:
musickatia [10]3 years ago
6 0

Answer:

umm no company can be run by kids because it's just too much for kids to handle

hope this helps

have a good day :)

Explanation:

GREYUIT [131]3 years ago
4 0

Answer:

i d k about that but I know it`s a Polish thing

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Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball
vlabodo [156]

Answer:

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt²......................... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²

Substitute into equation 1

S = 49(2) - 1/2(9.8)(2)²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m

6 0
3 years ago
Why does it take mars over 500 days to orbit the sun?
kakasveta [241]

Answer:

Mar's orbital path is more than that of Earth, thus it takes more number of days to orbit around the sun.

Explanation:

Mars takes over 500 days to orbit all the way around the sun than Earth because its distance from the sun (228 million kilometers) is greater than that of Earth (150 million kilometers) which takes it 365 days.

Planets that orbit closer to the sun take shorter time to orbit around the sun because the cover a shorter orbital distance and orbit faster than those planets further from the sun.

<u>For example</u>

Using Earth's distance from the sun, 150 million kilometers and the number of days taken to orbits the sun ,365 days and the distance Mars is from the Earth, 228 million kilometers, you can approximate the time Mar takes to orbit the sun as:

Earth 150 million kilometers  = 365 days

Mars  228 million kilometers= ?

Cross product ; (228 *365) /150 =555 -----(a value closer to that in the question)

6 0
3 years ago
Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He
umka2103 [35]
<h3>Answer:  130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

  • The initial speed (let's call that Vi) is 8.0 m/s
  • The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
  • This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

  • At 0 seconds, he's going 8 m/s
  • At the 1 second mark, he's slowing down to 8-2 = 6 m/s
  • At the 2 second mark, he's now at 6-2 = 4 m/s
  • At the 3 second mark, he's at 4-2 = 2 m/s
  • Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

  • F = force applied
  • m = mass
  • a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

8 0
3 years ago
If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out
Maru [420]

Answer:

power=work done÷time taken

2×5=10

10÷10=1

ans 1J per second

5 0
2 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
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