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DochEvi [55]
3 years ago
6

A student is helping her teacher move a 9.5 kg box of books. What net sideways force must she exert on the box to slide it acros

s the floor so that is accelerates at 1.0m/s2?
Physics
1 answer:
Lilit [14]3 years ago
6 0

As we know that mass of the box is given as

m = 9.5 kg

acceleration of the box is given as

a = 1 m/s^2

so by Newton's II law we can say

F = ma

now we can substitute all the values in it

F = 9.5 \times 1

F = 9.5 N

So it required 9.5 N force to slide it

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What elements are in NaOH
agasfer [191]

Answer: Sodim hydroxide have a formula of NaOH . So the elements present in the compound is sodium,oxygen and hydrogen.

Explanation:

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3 years ago
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What is the longest wavelength of radiation with enough energy to break carbon-carbon bonds?
iogann1982 [59]

The longest wavelength of radiation used to break carbon-carbon bonds is 344 nm.

<u>Explanation:</u>

The longest wavelength of radiation can also be stated as the minimum radiation frequency required to cut carbon-carbon bond should be equal to the threshold energy of the carbon-carbon bonds.

The threshold energy will be equal to the binding energy of the carbon-carbon bonds. As it is known that carbon-carbon bonds exhibit a binding energy of 348 kJ/mole, the threshold energy to break it, is determined as followed.

First, we have to convert the energy from kJ/mol to J, i.e., energy for the carbon-carbon molecules,

\text { Energy } = \frac{348 \mathrm{KJ} / \mathrm{mol}}{6.023 \times 10^{23} \text { photons }} \times 1 \text { mole } \times 1000 = 57.77 \times 10^{-20} = 5.78 \times 10^{-19} J

As,

         E=h v=\frac{h c}{\lambda}

So,

\lambda=\frac{h c}{E}=\frac{6.626 \times 10^{-34} \times 3 * 10^{8}}{5.78 \times 10^{-19}}=3.44 \times 10^{-7}

Thus, \lambda=344 \mathrm{nm} is the longest wavelength of radiation used to break carbon-carbon bonds.

5 0
3 years ago
A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle
Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

    The force which the massless bar is subjected to F = 22.5 \  N

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

     I_a =  I_G + M* r^2

Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

=>     \tau =  22.5 *  cos (47)

=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

4 0
3 years ago
Which statement about the properties of matter is true? *25*
mojhsa [17]

Answer:

A possible answer would be that chemical properties depend on phisical properties if and if only the phisical properties depend on the chemical ones ( see the laws of thermodinamics)

3 0
3 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
3 years ago
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