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3 years ago

Which option is an element?

2 answers:

7 0

The correct answer is carbon and the only element listed. Carbon Dioxide is made up of Carbon and Oxygen x2. This would be a compound. Air is made of water and another mixture of Hydrogen x2 and Oxygen.

Hope this helped! :))

Marta_Voda [28]3 years ago

3 0

Carbon is an element so answer choice B.

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A 15.4 kg block is dragged over a rough, horizontal surface by a constant force of 182 N acting at an angle of 24◦ above the hor

Dima020 [189]

Answer:

Explanation:

Considering Work done by friction is asked

Given

mass of block $m=15.4\ kg$

force $F=182\ N$

inclination $\theta =24^{\circ}$

block is displaced by $s=57.6\ m$

coefficient of kinetic friction $\mu _k=0.135$

Friction force $F_r=\mu _kN$

Normal reaction $N=mg-F\sin \theta =15.4\times 9.8-182\sin (24)$

$N=76.9\ N$

Friction Force $F_r=0.135\times 76.9=10.38\ N$

Work done by friction force

$W=f_r\cdot s$

$W=10.38\times 57.6=597.92\ N$

3 0

2 years ago

An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi

likoan [24]

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as: $\overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\$

In respect to the magnetic field; $\overrightarrow{B}=(0.027 i - 0.15 j) [T]$

The magnetic force can be written as;

$\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]$

Bear in mind $q =-1.6021x10^{-19} [C]$

thus,

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.

6 0

3 years ago

Is the following sentence true or false? Two atoms of the same element can have different numbers of protons.

pickupchik [31]

Two atoms of the same element can have different numbers of protons

false

diff nos neutrons ... isotopes,

diff protons ... diff elements

4 0

3 years ago

A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down

KIM [24]

<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>

6 0

3 years ago

Sphere A has a charge of +2 x 10 to the power of -6 coulomb and is brought into contact with a similar sphere, B which has a cha

Setler [38]

When sphere A and B are brought in contact and separated, charge on each sphere becomes [2x10^-6 + (-4x10^-6)]/ 2 = -1x10^-6 C.

That is, charge is equally separated and is the average of charges on both spheres. The reason behind equal charge on both spheres after separation is, when they are kept in contact, their potential difference becomes same.

7 0

3 years ago