Ask question

Ask question

All categories

3 years ago

5

Light travels 300,000,000 m/s, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in

one year.
State the speed in m/s, using scientific notation.

1 answer:

Julli [10]3 years ago

5 0

Explanation:

The speed of light is 300,000,000 m/s. There are 8 zeros, so in scientific notation, the speed is 3×10⁸ m/s.

You might be interested in

g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi

Arisa [49]

Answer:

$F_x = -\frac{6 C_6}{2^7}$

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by $\frac{C_6}{X_6}$

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = $\frac{C_6}{X_6}$

Force exerted by one atom upon another

$F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})$

or

$F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})$

or

$F_x = -\frac{6 C_6}{2^7}$

As we can see that the $C_6$ comes in positive and constant which represents that the force is negative that means the force is attractive in nature

5 0

3 years ago

How rolling friction is less than sliding friction?

vovangra [49]

Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.

6 0

3 years ago

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$

making $v_A$ the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

$v_A= 4m/s$

6 0

3 years ago

The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength

VARVARA [1.3K]

So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6

8 0

3 years ago

Charina says that when waves interact with an object, they will interfere with the object, and when waves interact with other wa

Fofino [41]

I would not agree with her since reflection and refraction happens only when waves hit an object. When, waves meet it is either it experiences constructive or destructive interference. Hope this answers the question. Have a nice day.

3 0

2 years ago

Read 2 more answers