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lisov135 [29]
2 years ago
5

How do I solve this problem

Physics
1 answer:
PilotLPTM [1.2K]2 years ago
5 0

Answer:

it is light

Explanation:

the arrow that says light is on the glass it must be near from tungsten

You might be interested in
Erwin Schrödinger developed the quantum model of the atom. What scientific knowledge was necessary for Schrödinger’s work?
Vikki [24]

The scientific knowledge was necessary for Schrödinger’s work is the discoverd of electron.

<h3>What is electron and how was it discovered?</h3>

Thomson carried out new experiments that led him to conclude that cathode rays were formed by particles that have a negative charge. Later, Thomson proved that these rays were deflected by applying an electric field. Thus, these particles were called electrons.

The discovery of the electron allowed Schrodinger to explain quantum models that are based on how electrons behave.

See more about quantum model at brainly.com/question/13352063

#SPJ1

6 0
1 year ago
A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coeff
natulia [17]

Answer:

a

H  =212.6 \  J

b

v  =  7.647  \  m/s

Explanation:

From the question we are told that

   The child's weight is  W_c  =  287 \ N

    The length of the sliding surface of the playground is  L =  7.20 \  m

    The coefficient of friction is  \mu =  0.120

      The angle is \theta = 31.0 ^o

      The initial  speed is  u =  0.559 \  m/s

Generally the normal force acting on the child is mathematically represented as

=>    N  =  mg  *  cos \theta

Note  m *  g  =  W_c

Generally the frictional force between the slide and the child is    

         F_f  =  \mu *  mg  *  cos \theta

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      F =m* g sin(\theta) - F_f

Here  F is the resultant force and it is represented as  F =  ma

=>   ma =   m* g sin(31.0)  - \mu *  mg  *  cos (31.0)

=>   a =  g sin(31.0)-  \mu *  g  *  cos (31.0)

=>  a =    9.8 *  sin(31.0) - 0.120 *  9.8  *  cos (31.0)

=>a =  4.039 \ m/s^2

So

   F_f  =  0.120  * 287  *  cos (31.0)

=> F_f  = 29.52 \  N

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     H  =  F_f  * L

=>  H  = 29.52 *  7.2

=>  H  =212.6 \  J

Generally from kinematic equation we have that

    v^2  =  u^2  +  2as

=>  v^2  =  0.559^2  +  2 * 4.039 * 7.2

=>  v  =  \sqrt{0.559^2  +  2 * 4.039 * 7.2}

=>  v  =  7.647  \  m/s

   

6 0
3 years ago
A stone is thrown vertically upwards with a speed of 30.0 m/s.
matrenka [14]

a)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s


b)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m



6 0
3 years ago
How much force should a mother exert to lift a 5.0-kg child​
andrew11 [14]

Answer:

F = 49N

Explanation:

F = mg

F = 5kg × 9.8m/s^2

F = 49kg.m/s^2 = 49N

8 0
3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
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