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olganol [36]
3 years ago
15

a given box of 2 kg is in contact with the floor experiences an applied force of 150 form somebody pushing it.while in motion,it

also experiences a frictional force of 7n in the opposite direction​. Calculate the weight of the box
Physics
1 answer:
ladessa [460]3 years ago
6 0

Weight = (mass) x (gravity)

Weight = (2kg) x (9.8 m/s^2)

Weight = 19.6 N

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A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
An integrated circuit is a _____.
Mademuasel [1]

closed and unsecure circuit


8 0
3 years ago
Read 2 more answers
#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

V=Ax

where A is the given area, so

4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

8 0
3 years ago
Velocity can best be described by which of the following statements?
pogonyaev

Answer: Velocity can best be described as, the speed in a given direction.

Explanation: To find the answer, we need to know more about the Velocity of a body.

<h3>What is Velocity of a body?</h3>
  • Velocity is the rate of change of displacement.
  • It's a vector quantity and is measured in m/s.
  • It can be positive, negative or zero.
  • A body is said to be in uniform motion, then its velocity remains constant.
  • Change in velocity can be a change in speed.
  • The magnitude of velocity is less than or equal to speed.

Thus, we can conclude that, the option C is best describing velocity.

Learn more about velocity here:

brainly.com/question/28108466

#SPJ4

8 0
2 years ago
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Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.
alina1380 [7]

Answer:

The time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

Explanation:

Given the intensity of the energy is 1.36\ kW/m^2

And the arriving energy is 3.55\times 10^9\ J

Also, the area in which energy is being arriving is 4.25\ m^2

Now, we will use relation between energy (E), intensity of energy (p), area (A) and time (T).

Where energy is in Joule, intensity is in kW/m^2, area is in m^2 and time is in seconds.

The equation is

E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds

So, the time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

4 0
3 years ago
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