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Studentka2010 [4]
1 year ago
5

Someone solvw this for me please

Physics
1 answer:
soldi70 [24.7K]1 year ago
6 0

The velocity of the mass at time, t = 1 s, is determined as 20 m/s.

<h3>Magnitude of the force experienced by the body</h3>

The magnitude of the force experienced by the body is calculated as follows;

|F| = √(8² + 6²)

|F| = 10 N

<h3>Acceleration of the mass</h3>

From Newton's second law of motion;

F = ma

where;

  • m is mass
  • a is acceleration

a = F/m

a = 10 /0.5

a = 20 m/s²

<h3>Velocity of the mass after 1 second</h3>

v = at

v = 20 x 1

v = 20 m/s

Learn more about velocity here: brainly.com/question/6504879

#SPJ1

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A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 5.5 kg box sliding across
Aloiza [94]

Answer: The box was moving with a velocity of 0.256m/s when it hit the spring

Explanation: Please see the attachments below

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3 years ago
A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r
Firlakuza [10]

Answer:

Speed of the this part is given as

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

P_1 + P_2 + P_3 = P_i

here we know that

P_1 = P_2

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

P_1 + P_2 = \sqrt{P^2 + P^2}

P_1 + P_2 = \sqrt2 mv

now from above equation we have

P_3 = -(P_1 + P_2)

mv_3 = -(\sqrt 2 mv)

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Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0
3 years ago
Help me please!!!
lilavasa [31]
Both planets are similar in shape and have a rocky surface. Not sure about the phases though
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At what position or positions on the x-axis is the electric field zero?
ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

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3 years ago
Choose 3 ways to keep our bodies healthy
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