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3 years ago

A 0.141 kg pinewood derby car is moving 1.33 m/s . What is its momentum?

Physics

1 answer:

kompoz [17]3 years ago

7 0

momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)

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how do u calculate the kinetic energy of a ball of mass 0.25kg being kicked vertically upwards with a speed of 5m/s

vfiekz [6]

Answer:

3.125J

Explanation:

K.E.= 1/2(mass)(velocity)^2

K.E.=1/2(0.25)(5)^2=3.125

6 0

3 years ago

Three forces act on a box on an inclined plane as shown in the diagram below. [Vectors

jok3333 [9.3K]

Answer:

zero

Explanation:

For the box remaining at rest, the total acceleration on the box must be zero. Since force F = m*a, the force F must also be zero.

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3 years ago

Consider the two vectors A = 3 î − ĵ and B = − î − 2 ĵ.

LUCKY_DIMON [66]

Answer:

(a) A+B = 2i-3j

(B) A-B = 4i + j

Explanation:

We have given two vectors A = 3i-j and B = -1-2j

We have to find the two vectors that is A+B and A-B

(A) In first art we have calculate A+B for this we have to add simply vector A and v ector B

So A+B = 3i-j-i-2j = 2i-3j

(B) In this part we have to find A-B for this we have to simply subtract B from A so A-B = 3i-j-(-i-2j) =3i-j+i+2j =4i+j

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3 years ago

When verifying Kirchhoff's rules, you measured: (a) temperature (b) resistance (c) time (d) current

Nadya [2.5K]

Answer:

option D

Explanation:

The correct answer is option D

When verifying Kirchhoff's law we measure current.

Kirchhoff's law deal with current and potential difference.

This law was first described by the German Scientist Gustav Kirchhoff.

Kirchhoff's law stated the current arriving at the node is equal to the current moving out of the node.

3 0

3 years ago

Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o

Anna35 [415]

Answer:

the correct one is 2. the equipotential lines must be closer together where the field has more intensity

Explanation:

The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.

In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by

V = k q / r

also the electric field and the electuary potential are related

E = $- \frac{dV}{dr}$

therefore the equipotential lines must be closer together where the field has more intensity

When checking the answers, the correct one is 2

3 0

3 years ago