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ASHA 777 [7]
3 years ago
7

Please help

Physics
1 answer:
stira [4]3 years ago
5 0

It’s D. The pitch would be lower. I’m 100% sure. I just took the test. I hope this helps you!! <3

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Una persona y su perro se encuentran inicialmente en reposo y paradas arriba de una caja que está apoyada sobre un piso rugoso.
alexandr1967 [171]

Answer:

La masa de la mascota es de 11,24 kg.

Explanation:

Dado que la masa total de = 80 kg

Tenemos

Fuerza de fricción = 80 kg × 9.81 × 0.3 = 235.44 N

La mascota salta horizontalmente con una velocidad, v, de 6.0 m / s para dar un movimiento de 1 m a la caja de masa y a la persona

Trabajo realizado por el salto de la mascota = (80 - m) × 9.81 × 0.3

La energía del movimiento del perro = 1/2 × m × v² = Trabajo realizado al saltar de la mascota

1/2 × m × 6² = (80 - m) × 9.81 × 0.3

20943 · m = 235440

m = 235440/20943 = 11,24 kg

5 0
3 years ago
On a Vernier Caliper, how do you know which mark to use on the very top scale?
madreJ [45]

<u>Answer</u>

To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use<em>. </em>


<u>Explanation</u>

A vernier caliper is an instrument that is used to measure the diameter of small circular objects such as diameter of a wires, thickness of an iron sheet.

The objects to be measured is place between the jaws of the calipers.

The vernier scale has two scales, the vernier scale and the main scale which is the very top scale.<em> To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use. </em>

4 0
3 years ago
The greater the mass is in an object, the higher resistance to a change in movement the object will have. Please select the best
Fofino [41]
This statement is true. The greater the mass is in an object, it is indeed the higher resistance to a change in movement the object will have. That only mean that the mass of an object and its resistance to change of movement is directly proportional.
3 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th
xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}

\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}

\displaystyle F'=9k\frac{q_1q_2}{d^2}

\displaystyle F'=9F

The force is now 9 times the original force

8 0
2 years ago
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