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swat32
3 years ago
11

A 0.2 kg plastic cart and a 20 kg lead cart both roll without friction on a horizontal surface. Equal forces are used to push bo

th carts forward for a time of 1 s, starting from rest. Is the momentum of the plastic cart greater than, less than, or equal to the momentum of the lead cart?
Physics
1 answer:
garik1379 [7]3 years ago
8 0

Answer:

The same

Explanation:

The change in momentum of each cart is equal to the impulse given by the force applied on the cart:

\Delta p = I = F \Delta t

where

F is the force applied

\Delta t is the time during which the force is applied

The force applied to both carts (F) is the same, as well as the time (\Delta t). This means that the change in momentum, \Delta p, is the same for both carts. But both carts start from rest (momentum = 0): this means that at the end, both carts will have exactly same momentum.

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Newtons law of motion for every action there’s an equal and opposite reaction.
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What are some tasks that organelles perform inside of a cell?
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Different organelles play different roles in the cell — for instance, mitochondria generate energy from food molecules; lysosomes break down and recycle organelles and macromolecules; and the endoplasmic reticulum helps build membranes and transport proteins throughout the cell.

Explanation:

Hopefully this helped!

4 0
3 years ago
A particle has a charge of -4.25 nC.
SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m

The distance from the electric field is 1.71436 m

4 0
3 years ago
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a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe
Vika [28.1K]

Answer:

<h2> 27m/s</h2>

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

5 0
3 years ago
An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto
kupik [55]

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

4 0
3 years ago
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