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castortr0y [4]
3 years ago
7

A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const

ant braking force, and are brought to rest. Which object travels the greater distance before stopping? a) the 4.0 kg object b) the 1.0 kg object c) Both objects travel the same distance d) It cannot be determined from the information given.
Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

0= u^2-2as_{1}

s_{1}=\dfrac{u^2}{2a}....(I)

Using newton law

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{1}=\dfrac{8}{F}

Now, For second object

Using equation of motion

v^2=u^2+2as

Put the value in the equation

0= u^2-2as_{2}

s_{2}=\dfrac{u^2}{2a}....(I)

Using newton law

F = ma

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{2}=\dfrac{8}{F}

Hence, Both objects travel the same distance.

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Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi
jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

n_{e} = number of electrons = 6.25 x 10⁶

q_{e} = charge on electron = - 1.6 x 10⁻¹⁹ C

n_{p} = number of protons = 7.75 x 10⁶

q_{p} = charge on proton =  1.6 x 10⁻¹⁹ C

Net charge is given as

Q = n_{e} q_{e} + n_{p} q_{p}

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0
3 years ago
What does muttered mean
Tema [17]
Muttered is the same thing like mumble. It's like when you are speaking and no one couldn't hear what you said. So Muttered is when you say something low .
3 0
3 years ago
Read 2 more answers
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krek1111 [17]

Explanation:

the table and the wooden block

6 0
3 years ago
Read 2 more answers
Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What
vesna_86 [32]

Answer:

1.02\cdot 10^{-8} N, repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have  same  sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

q_1=q_2=1.6\cdot 10^{-19}C is the magnitude of the two electrons

r=1.5\cdot 10^{-10} m is their separation

Substituting into the formula, we find the electric force between them:

F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N

And the force is repulsive, since the two electrons have same sign charge.

4 0
3 years ago
A proton moves from a location where V = 87 V to a spot where V = -40 V. (a) What is the change in the proton's kinetic energy?
Art [367]

Answer: a) 127 eV; b) there is no change of kinetic energy.

Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field.  This means the electric force do work in this trayectory and then the protons increased changes its speed.

If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then  they can not move to higher potential  if any  external force does work on the system.

In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons  can not move to lower potential region (V=-40V).

6 0
3 years ago
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