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DiKsa [7]
2 years ago
14

Large and small numbers are often best entered in exponential notation. For example, the mass of the earth is 5.98x1024 kg and t

he mass of an electron is 9.11x10-31 kg. Canvas does not have the capability to handle scientific notation. We will include the power of 10 with the units needed in the answer. The charge of an electron is -1.60x10-19 C. Determine the charge to mass ratio of the electron (in 1011 C/kg). (Hint: divide the electron's charge by its mass; the answer is a negative number with a positive exponent)
Physics
2 answers:
stich3 [128]2 years ago
5 0

Answer:

The charge to mass ratio is

m/e= -0.17*10^12kg/c

Explanation:

Step one :

Given data

mass of an electron m= 9.11x10-31kg

charge of electron q= -1.60x10-19C

Applying the equation

mv²/r=Bev

Where m =mass of charge

v= velocity of charge

r= distance

B=magnetic field

e= charge

Rearranging to get the charge-mass ratio we have

mv²/r=Bev

m/e=Br/v

(-1.60x10-19)/(9.11x10-31kg)=Br/v

-0.17*10^12

m/e= -0.17*10^12kg/c

Doss [256]2 years ago
4 0

Answer:

The charge to mass ratio is -1.76\times10^{11}

Explanation:

Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\

We need to find how much charge is contained  in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}

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Construction of a solar power plant is proposed for a desert area near a school. A student has hypothesized that the shade cast
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4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period

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In order to test the hypothesis the student must record the direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period. If the plants receive sunlight less than the above amount the plants should start dying. If not then the hypothesis is false.

Hence, the answer is 4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period.

4 0
2 years ago
Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.
solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination \theta =58.7^{\circ}C

Horizontal distance travel by Particle d=1200 m

Vertical height h=780 m

Let u be the initial velocity

calculating vertical distance

y=u\sin \theta +\frac{at^2}{2}

y=u\sin \theta t-\frac{gt^2}{2}-------1

Calculating horizontal distance

x=u\cos \theta \times t+0

t=\frac{x}{u\cos \theta }

put value of t in equation 1

y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2

y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }

\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y

u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}

u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}

at y=-780\ m\ x=1200 m

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6 0
2 years ago
A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release
Scilla [17]

Answer:

(A) As it moves farther and farther from Q, its speed will keep increasing.

Explanation:

When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.

Mathematically:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}

where:

r = distance between the charges

\epsilon_0= permittivity of free space

By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as  the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.

7 0
2 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

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Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
2 years ago
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