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3 years ago

Hi i have homework can you help me

Physics

1 answer:

soldier1979 [14.2K]3 years ago

6 0

Explanation:

The particle will be at rest when its velocity $v(t)$ is equal to zero. Recall that the velocity is simply the derivative of the position $x(t)$ with respect to time:

$v(t) = \dfrac{dx}{dt}$

Since $x(t) = t^2 - 2t + 2$

then

$v(t) = \dfrac{d}{dt}(t^2 - 2t + 2) = 2t - 2 = 0$

Solving for t, we find that the particle will be at rest at

$t = 1\:\text{s}$

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An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

3 years ago

two toy cars can roll without friction on a horizontal surface. Car A is a 5 kg. Car B is 0.5 kg. Both cars start from rest. Equ

Llana [10]

Answer:

A. The momentum of car A(5kg) is EQUAL to that of car B(0.5)

Explanation:

The moment, or impulse formula of the same forces acting on both car within 1 second is

$\Delta p = F * \Delta t$

In our case the forces are the same, the time duration of force acting on the cars are the same. Therefore, their momentum right after the force must also be the same.

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3 years ago

During a chemical reaction, Bromine (Br) would be expected to

Andrew [12]

Answer:

During a chemical reaction, Bromine (Br) would be expected to gain 1 valence electron to have a full octet.

Explanation:

In the periodic table the elements are ordered so that those with similar chemical properties are located close to each other.

The elements are arranged in horizontal rows, called periods, which coincide with the last electronic layer of the element. That is, an element with five electronic shells will be in the fifth period.

The columns of the table are called groups. The elements that make up each group coincide in their electronic configuration of valence electrons, that is, they have the same number of electrons in their last.

The elements tend to resemble the closest noble gases in terms of their electronic configuration of the last layer, that is, having eight electrons in the last layer to be stable.

Bromine belongs to group 17 (VII A), which indicates that it has 7 electrons in its last shell. So bromine requires more energy to lose all 7 electrons and generate stability, than it does to gain 1 electron and fill in 8 electrons to be stable. So:

During a chemical reaction, Bromine (Br) would be expected to gain 1 valence electron to have a full octet.

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2 years ago

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3 years ago

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Komok [63]

Answer:

Explanation:

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3 years ago