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IgorC [24]
2 years ago
6

As the electric potential near the ground increases during a thunderstorm, a positive charge current can move up pointed objects

, such as masts of ships, producing a luminous halo or glow known as ____.
Physics
1 answer:
cestrela7 [59]2 years ago
5 0

Answer:

St. Elmo's Fire - well known to old time sailors

You might be interested in
A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction betw
umka2103 [35]

Answer:

<h3>0.69</h3>

Explanation:

Using the Newtons law of motion;

\sum Fx = ma_x\\Fm - Ff = ma_x

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69

Hence the coefficient of kinetic friction between the box and floor is 0.69

7 0
3 years ago
Which is the best description of inertia? Inertia is
valentinak56 [21]

Answer:

The best description of inertia is an objects velocity

8 0
4 years ago
Consider matrices in R 3×5 . What is the maximum possible number of linearly independent column vectors?
raketka [301]

Answer:

if a matrix 3 x 5 it does not imply that the column vector are linearly independent.

Explanation:

A 3 x 5 matrix represents a linear map R^5 to R^3. we have 5 columns vectors of R^3. they can never be linearly independent because the dimension of R^3 is 3.

5 0
3 years ago
A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with
Lena [83]
REFER TO THE IMAGES for the SOLUTIONS TO YOUR PROBLEM. Each step will be explained here.

When you solve for velocities before or after collision, you need to remember the law of conservation of moment which can be expressed through this formula:

  BEFORE                   AFTER
m1v1+m2v2      =     m1v1 + m2v2

This basically means, the sum of momentum of 2 objects BEFORE collision is equal to the same 2 objects AFTER collision. 

The type of collision we have in your case is a 2D collision, where there is a gliding collision or they touch at an angle. So when you solve these type of problems, you have to consider the x and y components of motion. It makes things easier if you make a table like in the image to sort out your components.


STEP 1: COMPUTE FOR MOMENTUM BEFORE COLLISION for each OBJECT involved.
To solve for momentum, the formula is mass x velocity or mv:

STEP 1a: Body A: The problem states that before collision Body A is moving along the positive X-axis so the velocity will be +4.5 m/s. Notice that the velocity of the y component is 0 m/s. This is because BODY A is moving along the x-axis, with no mention that it deviated from it. 

STEP 1b: Body B: Body B is at rest before collision, that is why it is not moving at all, which means both x and y components are equal to 0.

STEP 1c: Get the sum of all X components and the Sum of Y components.

STEP 2:   COMPUTE FOR MOMENTUM AFTER COLLISION for each OBJECT involved.

Step 2a: BODY A: Notice that we now have an angle. hence the cos and sin. We do this because we are breaking or decomposing the diagonal velocity into its x and y component. To get the x-component you get the cos of the angle and multiply it to the momentum of the diagonal or overall velocity. For y-component, instead of cos, you get the sin. 

Step 2b: BODY B: Here we have unknowns, which we will derive later on. In this step, just plug in what you know into the formula. 

Step 2c: We already know the x and y momentum of the objects BEFORE collision and the law of conversation of momentum states that the momentum AFTER is the same. With this total we can move onto the next step.

STEP 3: Solving for the X and y component of the velocity of BODY B AFTER collision.

Step 3a: Using the formula given in the image, we plug in what we know first. We know the momentum of the BODY A already, so we can put it into the equation. We also know the sum of both momenta and we put that into the equation too. Now all we do is derive the formula. DO NOT FORGET THAT WE ARE TO USE ONLY X COMPONENTS. 

Step 3b: is the same as the previous step but instead, we use Y COMPONENTS only. 

STEP 4: Combining X and Y components to get the resultant velocity:
For this step you need to remember the Pythagorean theorem. This is applied here because when you draw a free body diagram of the velocities, it creates a right triangle where :
the hypotenuse represents the final velocity
the opposite side represents the y-component and;
the adjacent represents the x-component.

Refer to the image for the solution.

STEP 5: Solving for the angle at which BODY B is moving:
For this step you need to remember SOH CAH TOA to find the angle at which BODY B is moving. You already have all the components you need, including the hypotheses. You can use any of the functions, and they should come up with the same approximation. 

FINAL ANSWER: BODY B was moving at 1.35 m/s, 21 degrees above the x-axis. 

4 0
4 years ago
Your friend (68 kg) is wearing frictionless roller skates and standing still. You throw at her a 3.6 kg pumpkin with a velocity
Fudgin [204]

Answer:

Approximately 0.48\; \rm m\cdot s^{-1}.

Explanation:

Momentum would be conserved since there's no friction on this friend, and all other forces on her are balanced. Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)}\end{aligned}.

Momentum p the product of mass m and velocity v. That is: p = m \, v.

The initial momentum of this friend is 0\; \rm kg \cdot m \cdot s^{-1} since she was initially not moving (an initial velocity of 0\; \rm m\cdot s^{-1}.)

The initial momentum of the pumpkin would be:

\begin{aligned}p &= m \, v \\ &= 3.6 \; \rm kg \times 9.5\; \rm m\cdot s^{-1} \\ &= 34.2\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

Therefore:

\begin{aligned}& \text{Resultant momentum of (friend and pumpkin)} \\ =\; & \text{Initial momentum of (friend)} \\ & + \text{Initial momentum of (pumpkin)} \\ =\; &0\; {\rm kg \cdot m \cdot s^{-1}} + 34.2\; {\rm kg \cdot m \cdot s^{-1}} \\ =\; & 34.2\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}.

Rearrange the equation p = m \, v to find an expression for velocity v given momentum and mass:

\displaystyle v = \frac{p}{m}.

Note that the "final momentum of friend and pumpkin" in the previous equation refers to the resultant velocity of the friend with the pumpkin in her hand. Thus, it would necessary to use the combined mass of the friend and the pumpkin (68\; {\rm kg} + 3.6 \; {\rm kg}) when calculating the resultant velocity:

\begin{aligned}& \text{Resultant velocity of (friend and pumpkin)} \\ =\; & \frac{\text{Resultant momentum of (friend and pumpkin)}}{\text{Mass of (friend and pumpkin)}} \\ =\; & \frac{34.2\; {\rm kg \cdot m \cdot s^{-1}}}{68\; {\rm kg} + 3.6\; {\rm kg}} \\ \approx \; & 0.48\; \rm m \cdot s^{-1}\end{aligned}.

6 0
3 years ago
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